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Quantum Mechanics 

2 nd term 2002 

Martin Plenio 
Imperial College 

Version January 28, 2002 

Office hours: Tuesdays 
llam-12noon and 5pm- 6 pin! 
Office: Blackett 622 
Available at: 


I Quantum Mechanics 5 

1 Mathematical Foundations 11 

1.1 The quantum mechanical state space 11 

1.2 The quantum mechanical state space 12 

1.2.1 From Polarized Light to Quantum Theory .... 12 

1.2.2 Complex vector spaces 20 

1.2.3 Basis and Dimension 23 

1.2.4 Scalar products and Norms on Vector Spaces ... 26 

1.2.5 Completeness and Hilbert spaces 36 

1.2.6 Dirac notation 39 

1.3 Linear Operators 40 

1.3.1 Definition in Dirac notation 41 

1.3.2 Adjoint and Hermitean Operators 43 

1.3.3 Eigenvectors, Eigenvalues and the Spectral The- 
orem 45 

1.3.4 Functions of Operators 51 

1.4 Operators with continuous spectrum 57 

1.4.1 The position operator 57 

1.4.2 The momentum operator 61 

1.4.3 The position representation of the momentum 
operator and the commutator between position 

and momentum 62 

2 Quantum Measurements 65 

2.1 The projection postulate 65 

2.2 Expectation value and variance 70 

2.3 Uncertainty Relations 71 



2.3.1 The trace of an operator 74 

2.4 The density operator 76 

2.5 Mixed states, Entanglement and the speed of light .... 82 

2.5.1 Quantum mechanics for many particles 82 

2.5.2 How to describe a subsystem of some large system? 86 

2.5.3 The speed of light and mixed states 90 

2.6 Generalized measurements 92 

3 Dynamics and Symmetries 93 

3.1 The Schrodinger Equation 93 

3.1.1 The Heisenberg picture 96 

3.2 Symmetries and Conservation Laws 98 

3.2.1 The concept of symmetry 98 

3.2.2 Translation Symmetry and momentum conserva- 
tion 103 

3.2.3 Rotation Symmetry and angular momentum con- 
servation 105 

3.3 General properties of angular momenta 108 

3.3.1 Rotations 108 

3.3.2 Group representations and angular momentum 
commutation relations 110 

3.3.3 Angular momentum eigenstates 113 

3.4 Addition of Angular Momenta 116 

3.4.1 Two angular momenta 119 

3.5 Local Gauge symmetries and Electrodynamics 123 

4 Approximation Methods 125 

4.1 Time-independent Perturbation Theory 126 

4.1.1 Non-degenerate perturbation theory 126 

4.1.2 Degenerate perturbation theory 129 

4.1.3 The van der Waals force 132 

4.1.4 The Helium atom 136 

4.2 Adiabatic Transformations and Geometric phases .... 137 

4.3 Variational Principle 137 

4.3.1 The Rayleigh-Ritz Method 137 

4.4 Time-dependent Perturbation Theory 141 

4.4.1 Interaction picture 143 


4.4.2 Dyson Series 145 

4.4.3 Transition probabilities 146 

II Quantum Information Processing 153 

5 Quantum Information Theory 155 

5.1 What is information? Bits and all that 158 

5.2 From classical information to quantum information. . . . 158 

5.3 Distinguishing quantum states and the no-cloning theorem. 158 

5.4 Quantum entanglement: From qubits to ebits 158 

5.5 Quantum state teleportation 158 

5.6 Quantum dense coding 158 

5.7 Local manipulation of quantum states 158 

5.8 Quantum cyptography 158 

5.9 Quantum computation 158 

5.10 Entanglement and Bell inequalities 160 

5.11 Quantum State Teleportation 166 

5.12 A basic description of teleportation 167 


Part I 
Quantum Mechanics 




This lecture will introduce quantum mechanics from a more abstract 
point of view than the first quantum mechanics course that you took 
your second year. 

What I would like to achieve with this course is for you to gain a 
deeper understanding of the structure of quantum mechanics and of 
some of its key points. As the structure is inevitably mathematical, I 
will need to talk about mathematics. I will not do this just for the sake 
of mathematics, but always with a the aim to understand physics. At 
the end of the course I would like you not only to be able to understand 
the basic structure of quantum mechanics, but also to be able to solve 
(calculate) quantum mechanical problems. In fact, I believe that the 
ability to calculate (finding the quantitative solution to a problem, or 
the correct proof of a theorem) is absolutely essential for reaching a 
real understanding of physics (although physical intuition is equally 
important). I would like to go so far as to state 

If you can't write it down, then you do not understand it! 

With 'writing it down' I mean expressing your statement mathemati- 
cally or being able to calculate the solution of a scheme that you pro- 
posed. This does not sound like a very profound truth but you would be 
surprised to see how many people actually believe that it is completely 
sufficient just to be able to talk about physics and that calculations are 
a menial task that only mediocre physicists undertake. Well, I can as- 
sure you that even the greatest physicists don't just sit down and await 
inspiration. Ideas only come after many wrong tries and whether a try 
is right or wrong can only be found out by checking it, i.e. by doing 
some sorts of calculation or a proof. The ability to do calculations is 
not something that one has or hasn't, but (except for some exceptional 
cases) has to be acquired by practice. This is one of the reasons why 
these lectures will be accompanied by problem sheets (Rapid Feedback 
System) and I really recommend to you that you try to solve them. It 
is quite clear that solving the problem sheets is one of the best ways to 
prepare for the exam. Sometimes I will add some extra problems to the 
problem sheets which are more tricky than usual. They usually intend 


to illuminate an advanced or tricky point for which I had no time in 
the lectures. 

The first part of these lectures will not be too unusual. The first 
chapter will be devoted to the mathematical description of the quan- 
tum mechanical state space, the Hilbert space, and of the description of 
physical observables. The measurement process will be investigated in 
the next chapter, and some of its implications will be discussed. In this 
chapter you will also learn the essential tools for studying entanglement, 
the stuff such weird things as quantum teleportation, quantum cryptog- 
raphy and quantum computation are made of. The third chapter will 
present the dynamics of quantum mechanical systems and highlight the 
importance of the concept of symmetry in physics and particularly in 
quantum mechanics. It will be shown how the momentum and angular 
momentum operators can be obtained as generators of the symmetry 
groups of translation and rotation. I will also introduce a different kind 
of symmetries which are called gauge symmetries. They allow us to 'de- 
rive' the existence of classical electrodynamics from a simple invariance 
principle. This idea has been pushed much further in the 1960's when 
people applied it to the theories of elementary particles, and were quite 
successful with it. In fact, t'Hooft and Veltman got a Nobel prize for it 
in 1999 work in this area. Time dependent problems, even more than 
time-independent problems, are difficult to solve exactly and therefore 
perturbation theoretical methods are of great importance. They will 
be explained in chapter 5 and examples will be given. 

Most of the ideas that you are going to learn in the first five chapters 
of these lectures are known since about 1930, which is quite some time 
ago. The second part of these lectures, however, I will devote to topics 
which are currently the object of intense research (they are also my 
main area of research). In this last chapter I will discuss topics such as 
entanglement, Bell inequalities, quantum state teleportation, quantum 
computation and quantum cryptography. How much of these I can 
cover depends on the amount of time that is left, but I will certainly talk 
about some of them. While most physicists (hopefully) know the basics 
of quantum mechanics (the first five chapters of these lectures), many 
of them will not be familiar with the content of the other chapters. So, 
after these lectures you can be sure to know about something that quite 
a few professors do not know themselves! I hope that this motivates 


you to stay with me until the end of the lectures. 

Before I begin, I would like to thank, Vincenzo Vitelli, John Pa- 
padimitrou and William Irvine who took this course previously and 
spotted errors and suggested improvements in the lecture notes and 
the course. These errors are fixed now, but I expect that there are 
more. If you find errors, please let me know (ideally via email so that 
the corrections do not get lost again) so I can get rid of them. 

Last but not least, I would like to encourage you both, to ask ques- 
tions during the lectures and to make use of my office hours. Questions 
are essential in the learning process, so they are good for you, but I also 
learn what you have not understood so well and help me to improve my 
lectures. Finally, it is more fun to lecture when there is some feedback 
from the audience. 


Chapter 1 

Mathematical Foundations 

Before I begin to introduce some basics of complex vector spaces and 
discuss the mathematical foundations of quantum mechanics, I would 
like to present a simple (seemingly classical) experiment from which we 
can derive quite a few quantum rules. 

1.1 The quantum mechanical state space 

When we talk about physics, we attempt to find a mathematical de- 
scription of the world. Of course, such a description cannot be justified 
from mathematical consistency alone, but has to agree with experimen- 
tal evidence. The mathematical concepts that are introduced are usu- 
ally motivated from our experience of nature. Concepts such as position 
and momentum or the state of a system are usually taken for granted 
in classical physics. However, many of these have to be subjected to 
a careful re-examination when we try to carry them over to quantum 
physics. One of the basic notions for the description of a physical sys- 
tem is that of its 'state'. The 'state' of a physical system essentially can 
then be defined, roughly, as the description of all the known (in fact 
one should say knowable) properties of that system and it therefore rep- 
resents your knowledge about this system. The set of all states forms 
what we usually call the state space. In classical mechanics for example 
this is the phase space (the variables are then position and momentum), 
which is a real vector space. For a classical point-particle moving in 




one dimension, this space is two dimensional, one dimension for posi- 
tion, one dimension for momentum. We expect, in fact you probably 
know this from your second year lecture, that the quantum mechanical 
state space differs from that of classical mechanics. One reason for this 
can be found in the ability of quantum systems to exist in coherent 
superpositions of states with complex amplitudes, other differences re- 
late to the description of multi-particle systems. This suggests, that a 
good choice for the quantum mechanical state space may be a complex 
vector space. 

Before I begin to investigate the mathematical foundations of quan- 
tum mechanics, I would like to present a simple example (including 
some live experiments) which motivates the choice of complex vector 
spaces as state spaces a bit more. Together with the hypothesis of 
the existence of photons it will allow us also to 'derive', or better, to 
make an educated guess for the projection postulate and the rules for 
the computation of measurement outcomes. It will also remind you 
of some of the features of quantum mechanics which you have already 
encountered in your second year course. 

1.2 The quantum mechanical state space 

In the next subsection I will briefly motivate that the quantum mechan- 
ical state space should be a complex vector space and also motivate 
some of the other postulates of quantum mechanics 

1.2.1 From Polarized Light to Quantum Theory 

Let us consider plane waves of light propagating along the z-axis. This 
light is described by the electric field vector E orthogonal on the di- 
rection of propagation. The electric field vector determines the state 
of light because in the cgs-system (which I use for convenience in this 
example so that I have as few e and /x as possible.) the magnetic field 

— * — * 

is given by B = e z x E. Given the electric and magnetic field, Maxwells 
equations determine the further time evolution of these fields. In the 
absence of charges, we know that E(r,t) cannot have a z-component, 



so that we can write 

E(f, t) = E x (f, t)e x + E y (f, t)e y 

'E x (f,ty 


The electric field is real valued quantity and the general solution of the 
free wave equation is given by 

E x (f, t) = E x cos(kz — out + a x ) 
E y (f, t) = Ey cos(kz — out + OLy) . 

Here k = 2n/\ is the wave-number, ou = litv the frequency, a x and a y 
are the real phases and E x and Ey the real valued amplitudes of the 
field components. The energy density of the field is given by 




(E"(r,t) + B 2 (f,t)) 

El) 2 cos 2 (kz -out + a x ) + (Ey) cos\kz - out + a y ) 

For a fixed position r*we are generally only really interested in the time- 
averaged energy density which, when multiplied with the speed of light, 
determines the rate at which energy flows in z-direction. Averaging over 
one period of the light we obtain the averaged energy density e(r) with 

^ = iz l(£°) 2 + (K) 




For practical purposes it is useful to introduce the complex field com 

E y (f,t) = Re{E y e^ kz -^) , 

E x (f,t) = Re(E x e 

i(kz-ujt) \ 


with E x = E^.e tax and E y = Eye tay . Comparing with Eq. (1.2) we find 
that the averaged energy density is given by 

e(r) = 


\E X \ 2 + \Ey 

Usually one works with the complex field 

E(f,t) = (E x e x + E y e y )e^ kz -^ = ( E A S kz ~^ 





This means that we are now characterizing the state of light by a vector 
with complex components. 

The polarization of light waves are described by E x and E y . In the 
general case of complex E x and E y we will have elliptically polarized 
light. There are a number of important special cases (see Figures 1.1 
for illustration). 

1. E y = 0: linear polarization along the x-axis. 

2. E x = 0: linear polarization along the y-axis. 

3. E x — E y : linear polarization along 45°-axis. 
A. E y — iE x : Right circularly polarized light. 
5. E y = —iE x : Left circularly polarized light. 

Figure 1.1: Left figure: Some possible linear polarizations of light, 
horizontally, vertically and 45 degrees. Right figure: Left- and right- 
circularly polarized light. The light is assumed to propagate away from 

In the following I would like to consider some simple experiments 
for which I will compute the outcomes using classical electrodynam- 
ics. Then I will go further and use the hypothesis of the existence of 
photons to derive a number of quantum mechanical rules from these 

Experiment I: Let us first consider a plane light wave propagating 
in z-direction that is falling onto an x-polarizer which allows x-polarized 



light to pass through (but not y polarized light). This is shown in 
figure 1.2. After passing the polarizer the light is x-polarized and from 

Figure 1.2: Light of arbitrary polarization is hitting a x-polarizer. 

the expression for the energy density Eq. (1.4) we find that the ratio 
between incoming intensity /«„ (energy density times speed of light) 
and outgoing intensity I out is given by 


\E X \ 


E x \ 

2 + 




So far this looks like an experiment in classical electrodynamics or 

Quantum Interpretation: Let us change the way of looking at 
this problem and thereby turn it into a quantum mechanical experi- 
ment. You have heard at various points in your physics course that light 
comes in little quanta known as photons. The first time this assumption 
had been made was by Planck in 1900 'as an act of desperation' to be 
able to derive the blackbody radiation spectrum. Indeed, you can also 
observe in direct experiments that the photon hypothesis makes sense. 
When you reduce the intensity of light that falls onto a photodetector, 
you will observe that the detector responds with individual clicks each 
triggered by the impact of a single photon (if the detector is sensitive 
enough). The photo-electric effect and various other experiments also 
confirm the existence of photons. So, in the low-intensity limit we have 
to consider light as consisting of indivisible units called photons. It is 
a fundamental property of photons that they cannot be split - there is 
no such thing as half a photon going through a polarizer for example. 
In this photon picture we have to conclude that sometimes a photon 
will be absorbed in the polarizer and sometimes it passes through. If 
the photon passes the polarizer, we have gained one piece of informa- 
tion, namely that the photon was able to pass the polarizer and that 
therefore it has to be polarized in x-direction. The probability p for 



the photon to pass through the polarizer is obviously the ratio between 
transmitted and incoming intensities, which is given by 

P= rEi 19. ~ i F i9 • 

\E X \ 



2 + 

\ E y 


If we write the state of the light with normalized intensity 

-» E x 

E N = =e x + =e y , (1.8) 

VKl 2 + |£ y | 2 ^p + S 2 

then in fact we find that the probability for the photon to pass the x- 
polarizer is just the square of the amplitude in front of the basis vector 
e x \ This is just one of the quantum mechanical rules that you have 
learned in your second year course. 

Furthermore we see that the state of the photon after it has passed 
the x-polarizer is given by 

E N = e x , (1.9) 

ie the state has has changed from {^ x ^j to (^)- This transformation 
of the state can be described by a matrix acting on vectors, ie 

E a 


The matrix that I have written here has eigenvalues and 1 and is 
therefore a projection operator which you have heard about in the sec- 
ond year course, in fact this reminds strongly of the projection postulate 
in quantum mechanics. 

Experiment II: Now let us make a somewhat more complicated 
experiment by placing a second polarizer behind the first x-polarizer. 
The second polarizer allows photons polarized in x' direction to pass 
through. If I slowly rotate the polarizer from the x direction to the y 
direction, we observe that the intensity of the light that passes through 
the polarizer decreases and vanishes when the directions of the two 
polarizers are orthogonal. I would like to describe this experiment 
mathematically. How do we compute the intensity after the polarizer 



now? To this end we need to see how we can express vectors in the 
basis chosen by the direction x' in terms of the old basis vectors e x ,e y . 

The new rotated basis e 1 ^, e! (see Fig. 1.3) can be expressed by the 
old basis by 

e! x = cos <f>e x + sin <f>e y e y = — sin <j>e x + cos (j>e y (1-H) 
and vice versa 

e x = cos (pe x — sm. (pe y e y — sin (f>e! x + cos . (1-12) 

Note that cos — e! x • e x and sin <fi = e* x ■ e y where I have used the real 
scalar product between vectors. 

Figure 1.3: The x'- basis is rotated by an angle with respect to the 
original x-basis. 

The state of the x-polarized light after the first polarizer can be 
rewritten in the new basis of the x'-polarizer. We find 

E = E x e x = E x cos 04 - E x sin^ej, = E X (2 X ■ e x )£ x - E x {g y ■ e y )£ y 

Now we can easily compute the ratio between the intensity before 
and after the x'-polarizer. We find that it is 

lafter i -» 1 2 

= 1^-41 =cos 2 (j) (1.13) 


or if we describe the light in terms of states with normalized intensity 
as in equation 1.8, then we find that 



— * 

where En is the normalized intensity state of the light after the x- 
polarizer. This demonstrates that the scalar product between vectors 
plays an important role in the calculation of the intensities (and there- 
fore the probabilities in the photon picture). 

Varying the angle between the two bases we can see that the ratio 
of the incoming and outgoing intensities decreases with increasing angle 
between the two axes until the angle reaches 90° degrees. 

Interpretation: Viewed in the photon picture this is a rather 
surprising result, as we would have thought that after passing the x- 
polarizer the photon is 'objectively' in the x-polarized state. However, 
upon probing it with an rr'-polarizer we find that it also has a quality of 
an x'-polarized state. In the next experiment we will see an even more 
worrying result. For the moment we note that the state of a photon can 
be written in different ways and this freedom corresponds to the fact 
that in quantum mechanics we can write the quantum state in many 
different ways as a quantum superpositions of basis vectors. 

Let us push this idea a bit further by using three polarizers in a 

Experiment III: If after passing the x-polarizer, the light falls onto 
a y-polarizer (see Fig 1.4), then no light will go through the polarizer 
because the two directions are perpendicular to each other. This sim- 

Figure 1.4: Light of arbitrary polarization is hitting a x-polarizer and 
subsequently a y-polarizer. No light goes through both polarizers. 

pie experimental result changes when we place an additional polarizer 
between the x and the y-polarizer. Assume that we place a x'-polarizer 
between the two polarizers. Then we will observe light after the y- 
polarizer (see Fig. 1.5) depending on the orientation of x'. The light 
after the last polarizer is described by E y e y . The amplitude E y is calcu- 
lated analogously as in Experiment II. Now let us describe the (x-x'-y) 
experiment mathematically. The complex electric field (without the 
time dependence) is given by 



Figure 1.5: An x'-polarizer is placed in between an x-polarizer and a 
y-polarizer. Now we observe light passing through the y-polarizer. 

before the x-polarizer: 

E± = E x e x + E y e y . 

after the x-polarizer: 

E 2 = {E x e x )e x = E x e x = E x cos 04 ~ E x sin0e^. 

after the x'-polarizer: 

E 3 = (E 2 e x )e x = E x cos <pe x = E x cos 2 4>e x + E x cos sin <pe y . 
after the y-polarizer: 

E4 = (E 3 e y )e y = E x cos(j)sm(f>ey = E y e y . 

Therefore the ratio between the intensity before the x'-polarizer and 
after the y-polarizer is given by 

^^ = cos 2 0sin 2 (1.15) 

■* before 

Interpretation: Again, if we interpret this result in the photon 
picture, then we arrive at the conclusion, that the probability for the 
photon to pass through both the x' and the y polarizer is given by 
cos 2 <f> sin 2 0. This experiment further highlights the fact that light of 
one polarization may be interpreted as a superposition of light of other 
polarizations. This superposition is represented by adding vectors with 
complex coefficients. If we consider this situation in the photon picture 
we have to accept that a photon of a particular polarization can also 
be interpreted as a superposition of different polarization states. 



Conclusion: All these observations suggest that complex vectors, 
their amplitudes, scalar products and linear transformations between 
complex vectors are the basic ingredient in the mathematical structure 
of quantum mechanics as opposed to the real vector space of classi- 
cal mechanics. Therefore the rest of this chapter will be devoted to a 
more detailed introduction to the structure of complex vector-spaces 
and their properties. 

Suggestions for further reading: 

G. Baym Lectures on Quantum Mechanics, W.A. Benjamin 1969. 
P.A.M. Dirac, The principles of Quantum Mechanics, Oxford Univer- 
sity Press 1958 

End of 1 st lecture 

1.2.2 Complex vector spaces 

I will now give you a formal definition of a complex vector space and 
will then present some of its properties. Before I come to this definition, 
I introduce a standard notation that I will use in this chapter. Given 
some set V we define 

1. \/\x) E V means: For all \x) that lie in V. 

2. 3\x) eV means: There exists an element \x) that lies in V. 

Note that I have used a somewhat unusual notation for vectors. I 
have replaced the vector arrow on top of the letter by a sort of bracket 
around the letter. I will use this notation when I talk about complex 
vectors, in particular when I talk about state vectors. 
Now I can state the definition of the complex vector space. It will 
look a bit abstract at the beginning, but you will soon get used to it, 
especially when you solve some problems with it. 

Definition 1 Given a quadruple (V, C, +, •) where V is a set of ob- 
jects (usually called vectors), C denotes the set of complex numbers, 



'+' denotes the group operation of addition and '■' denotes the mul- 
tiplication of a vector with a complex number. (V, C, +, •) is called a 
complex vector space if the following properties are satisfied: 

1. (V,+) is an Abelian group, which means that 


V|a), \b) G V \a) + \b) G V. 

( closure ) 


V|a>, |6>, |c> G V =► \a) + (|6> + \c)) = 

(|o> + |6» + |c>- 



3\0) eV so that V|a) e V =>■ |a) + \0) = \ 

a) . zero ) 


V|a) G V : 3(-|a)) G V so that \a) + (-\a)) 

— \0). (inverse) 


\/\a),\b) EV ^\a) + \b) = \b) + \a). 


2. The 

Scalar multiplication satisfies 


Va G C, \x) G V => a\x) G V 


y\x) eV l • \x) = \x) 



Vc, d e C,\x) eV (c • d) ■ \x) = c • (d • 

\x)) (associative) 


Vc, d G C, |y) G V c • (\x) + = c 

■\x) + c- \y) 

and (c + d) • \x) — c • \x) + c • \y) . 


This definition looks quite abstract but a few examples will make it 


1. A simple proof 

I would like to show how to prove the statement • \x) = \0). 
This might look trivial, but nevertheless we need to prove it, as it 
has not been stated as an axiom. From the axioms given in Def. 
1 we conclude. 

\0) (1 f -\x) + \x) 

{2 = b) -\x) + 1 • \x) 

= -\x) + (1 + 0) • \x) 

(2 = d) -\x) + l-\x)+0-\x) 




\x) + \x) + • \x) 

(1 f \O)+0-\x) 
^ 0-\x) . 

2. TheC 2 

This is the set of two-component vectors of the form 


where the a, are complex numbers. The addition and scalar mul- 
tiplication are defined as 

""Ik) (1-17) 
a 2 + b 2 J 


It is now easy to check that V = C 2 together with the addition 
and scalar multiplication defined above satisfy the definition of 
a complex vector space. (You should check this yourself to get 
some practise with the axioms of vector space.) The vector space 
C 2 is the one that is used for the description of spin-| particles 
such as electrons. 

The set of real functions of one variable f : R — > R 
The group operations are defined as 

(/i + /2)(x) := fi(x) + f 2 (x) 
(c-f)(x) := c-f(x) 

Again it is easy to check that all the properties of a complex 
vector space are satisfied. 

Complex n x n matrices 
The elements of the vector space are 

1 mn • • • min N 
M= j •-. j , (1.19) 

\ m nl . . . m nn J 



where the are arbitrary complex numbers. The addition and 
scalar multiplication are defined as 

(I In \ ( !>\. 


a nn ) 

bin \ 

■ b r , 

( a u + 6n 

Oln + bi n \ 

ain \ 

■ a r , 

\ &nl +b n l ■ ■ ■ a nn + b nn j 

( c • an ... c- a in ^ 
\ c ■ a n i ... c • a nn ) 

Again it is easy to confirm that the set of complex nxn matrices 
with the rules that we have defined here forms a vector space. 
Note that we are used to consider matrices as objects acting on 
vectors, but as we can see here we can also consider them as 
elements (vectors) of a vector space themselves. 

Why did I make such an abstract definition of a vector space? Well, 
it may seem a bit tedious, but it has a real advantage. Once we have 
introduced the abstract notion of complex vector space anything we 
can prove directly from these abstract laws in Definition 1 will hold 
true for any vector space irrespective of how complicated it will look 
superficially. What we have done, is to isolate the basic structure of 
vector spaces without referring to any particular representation of the 
elements of the vector space. This is very useful, because we do not 
need to go along every time and prove the same property again when 
we investigate some new objects. What we only need to do is to prove 
that our new objects have an addition and a scalar multiplication that 
satisfy the conditions stated in Definition 1. 

In the following subsections we will continue our exploration of the 
idea of complex vector spaces and we will learn a few useful properties 
that will be helpful for the future. 

1.2.3 Basis and Dimension 

Some of the most basic concepts of vector spaces are those of linear 
independence, dimension and basis. They will help us to express vec- 



tors in terms of other vectors and are useful when we want to define 
operators on vector spaces which will describe observable quantities. 

Quite obviously some vectors can be expressed by linear combina- 
tions of others. For example 


It is natural to consider a given set of vectors {\x)\, . . . ,\x)k} and to 
ask the question, whether a vector in this set can be expressed as a 
linear combination of the others. Instead of answering this question 
directly we will first consider a slightly different question. Given a set 
of vectors {\x)\, . . . ,\x)k}, can the null vector \0) can be expressed as 
a linear combination of these vectors? This means that we are looking 
for a linear combination of vectors of the form 

\ 1 \x) 1 + ... + \ 2 \x) k = \0) . (1.21) 

Clearly Eq. (1.21) can be satisfied when all the \ vanish. But this case 
is trivial and we would like to exclude it. Now there are two possible 
cases left: 

a) There is no combination of Aj's, not all of which are zero, that 
satisfies Eq. (1.21). 

b) There are combinations of Aj's, not all of which are zero, that 
satisfy Eq. (1.21). 

These two situations will get different names and are worth the 

Definition 2 A set of vectors {\x} 1 , . . . , \x) k } is called linearly inde- 
pendent if the equation 

X 1 \x) 1 + ... + X 2 \x) k = \0) (1.22) 

has only the trivial solution X± — . . . — A& = 0. 

If there is a nontrivial solution to Eq. (1.22), i.e. at least one of the 
Aj 7^ ; then we call the vectors {\x) l) . . . , \x) k } linearly dependent. 

Now we are coming back to our original question as to whether 
there are vectors in {\x)i, . . . , \x)k} that can be expressed by all the 
other vectors in that set. As a result of this definition we can see the 



Lemma 3 For a set o/linearly independent vectors {\x) 1 , . . . , \x) k }, 
no \x) i can be expressed as a linear combination of the other vectors, 
i.e. one cannot find Xj that satisfy the equation 

X 1 \x) 1 + ... + X i - 1 \x) i _ 1 + X i+1 \x) i+1 + ... + X k \x) k = \x) i . (1.23) 

In a set o/ linearly dependent vectors {\x) 1 , . . . , \x) k } there is at least 
one \x) i that can be expressed as a linear combination of all the other 

Proof: Exercise! □ 

Example: The set {|C)} consisting of the null vector only, is linearly 

In a sense that will become clearer when we really talk about quan- 
tum mechanics, in a set of linearly independent set of vectors, each 
vector has some quality that none of the other vectors have. 

After we have introduced the notion of linear dependence, we can 
now proceed to define the dimension of a vector space. I am sure that 
you have a clear intuitive picture of the notion of dimension. Evidently 
a plain surface is 2-dimensional and space is 3-dimensional. Why do 
we say this? Consider a plane, for example. Clearly, every vector in 
the plane can be expressed as a linear combination of any two linearly 
independent vectors |e)i, Je)2- As a result you will not be able to find a 
set of three linearly independent vectors in a plane, while two linearly 
independent vectors can be found. This is the reason to call a plane a 
two-dimensional space. Let's formalize this observation in 

Definition 4 The dimension of a vector space V is the largest number 
of linearly independent vectors in V that one can find. 

Now we introduce the notion of basis of vector spaces. 

Definition 5 A set of vectors {\x) v . . . , \x) k } is called a basis of a 
vector space V if 



a) \x) 1 , . . . , \x) k are linearly independent. 

b) V\ x ) e V : 3A, € C x = E-=i A^x).. 

Condition b) states that it is possible to write every vector as a 
linear combination of the basis vectors. The first condition makes sure 
that the set {\x) 1 , . . . , \x) k } is the smallest possible set to allow for con- 
dition b) to be satisfied. It turns out that any basis of an A-dimensional 
vector space V contains exactly N vectors. Let us illustrate the notion 
of basis. 


1. Consider the space of vectors C 2 with two components. Then the 
two vectors 


form a basis of C 2 . A basis for the C N can easily be constructed 
in the same way. 

2. An example for an infinite dimensional vector space is the space 
of complex polynomials, i.e. the set 

V = {c + dz + . . . + c fc 2 fc | arbitrary k and Vq G C} . (1.25) 

Two polynomials are equal when they give the same values for all 
z G C. Addition and scalar multiplication are defined coefficient 
wise. It is easy to see that the set {1, z, z 2 , . . .} is linearly inde- 
pendent and that it contains infinitely many elements. Together 
with other examples you will prove (in the problem sheets) that 
Eq. (1.25) indeed describes a vector space. 

End of 2 nd lecture 

1.2.4 Scalar products and Norms on Vector Spaces 

In the preceding section we have learnt about the concept of a basis. 
Any set of linearly independent vectors of an N dimensional vector 



space V form a basis. But not all such choices are equally convenient. 
To find useful ways to chose a basis and to find a systematic method 
to find the linear combinations of basis vectors that give any arbitrary 
vector \x) G V we will now introduce the concept of scalar product 
between two vectors. This is not to be confused with scalar multiplica- 
tion which deals with a complex number and a vector. The concept of 
scalar product then allows us to formulate what we mean by orthogo- 
nality. Subsequently we will define the norm of a vector, which is the 
abstract formulation of what we normally call a length. This will then 
allow us to introduce orthonormal bases which are particularly handy. 

The scalar product will play an extremely important role in quan- 
tum mechanics as it will in a sense quantify how similar two vectors 
(quantum states) are. In Fig. 1.6 you can easily see qualitatively that 
the pairs of vectors become more and more different from left to right. 
The scalar product puts this into a quantitative form. This is the rea- 
son why it can then be used in quantum mechanics to quantify how 
likely it is for two quantum states to exhibit the same behaviour in an 

Figure 1.6: The two vectors on the left are equal, the next pair is 
'almost' equal and the final pair is quite different. The notion of equal 
and different will be quantified by the scalar product. 

To introduce the scalar product we begin with an abstract formu- 
lation of the properties that we would like any scalar product to have. 
Then we will have a look at examples which are of interest for quantum 

Definition 6 A complex scalar product on a vector space assigns to 
any two vectors \x), \y) G V a complex number (\x), \y)) G C satisfying 
the following rules 



1. y\x),\y),\z)eV,a i eC: 

(\x),ai\y) + a 2 \z)) = ai(\x), \y)) + a 2 (\x), \z)) (linearity) 

2. V|ar), \y) E V : (\x), \y)) = (\y), \x))* (symmetry) 

3. y\x) e V : (\x), \x)) > (positivity) 

4. V\x)eV:(\x),\x)) = 0<*\x) = \O) 

These properties are very much like the ones that you know from 
the ordinary dot product for real vectors, except for property 2 which 
we had to introduce in order to deal with the fact that we are now 
using complex numbers. In fact, you should show as an exercise that 
the ordinary condition (\x),y) = (\y), \x)) would lead to a contradiction 
with the other axioms if we have complex vector spaces. Note that we 
only defined linearity in the second argument. This is in fact all we 
need to do. As an exercise you may prove that any scalar product is 
anti-linear in the first component, i.e. 

V|x), \y), \z)EV,aeC: (a\x)+P\y), \z)) = a*(\x), \ z )) + (3*(\y), \z)) . 


Note that vector spaces on which we have defined a scalar product are 
also called unitary vector spaces. To make you more comfortable 
with the scalar product, I will now present some examples that play 
significant roles in quantum mechanics. 


1. The scalar product in C n . 

Given two complex vectors \x), \y) G C" with components Xi and 
yi we define the scalar product 

(l*>,lv» = i>.-!/i ( L27 ) 

where * denotes the complex conjugation. It is easy to convince 
yourself that Eq. (1.27) indeed defines a scalar product. (Do it!). 



2. Scalar product on continuous square integrable functions 
A square integrable function ip e £ 2 (R) is one that satisfies 

Eq. (1.28) already implies how to define the scalar product for 
these square integrable functions. For any two functions ip, <f) G 
£ 2 (R) we define 

Again you can check that definition Eq. (1.29) satisfies all prop- 
erties of a scalar product. (Do it!) We can even define the scalar 
product for discontinuous square integrable functions, but then 
we need to be careful when we are trying to prove property 4 for 
scalar products. One reason is that there are functions which are 
nonzero only in isolated points (such functions are discontinuous) 
and for which Eq. (1.28) vanishes. An example is the function 

The solution to this problem lies in a redefinition of the elements 
of our set. If we identify all functions that differ from each other 
only in countably many points then we have to say that they are 
in fact the same element of the set. If we use this redefinition then 
we can see that also condition 4 of a scalar product is satisfied. 

An extremely important property of the scalar product is the Schwarz 
inequality which is used in many proofs. In particular I will used it 
to prove the triangular inequality for the length of a vector and in the 
proof of the uncertainty principle for arbitrary observables. 

Theorem 7 (The Schwarz inequality) For any \x), \y) eV we have 



1 for x = 

anywhere else 




Proof: For any complex number a we have 

< (\x)+a\y),\x)+a\y)) 

= (\x), \x)) + a(\x), \y)) + a*(\y), \x)) + \a\ 2 (\y), \y)) 

= (\x), \x)) + 2vRe(\x), \y)) - 2wlm(\x), \y)) + (v 2 + w 2 )(\y), 

=■ f(v,w) (1.31) 

In the definition of f(v,w) in the last row we have assumed a = v + 
iw. To obtain the sharpest possible bound in Eq. (1.30), we need to 
minimize the right hand side of Eq. (1.31). To this end we calculate 


= ^(v,w) = 2Re(\x),\y)) + 2v(\y),\y)) (1.32) 

= ^(v,w) = -2Im(\x),\y)) + 2w(\y),\y)) . (1.33) 

Solving these equations, we find 

Re(\x),\y))-Hm(\x),\y)) (\y),\x)) 
a mm -v mm + iw mm - {ly)Ay)) - {{y)Ay)) . 


Because all the matrix of second derivatives is positive definite, we 
really have a minimum. If we insert this value into Eq. (1.31) we 

0< (WlW )-MJggM) (1.36) 

This implies then Eq. (1.30). Note that we have equality exactly if the 
two vectors are linearly dependent, i.e. if \x) = j\y) □ . 

Quite a few proofs in books little bit shorter than this one 

because they just use Eq. (1.34) and do not justify its origin as it was 
done here. 

Having defined the scalar product, we are now in a position to define 
what we mean by orthogonal vectors. 

Definition 8 Two vectors \x), \y) e V are called orthogonal if 

(|*),|l/» = • (1-36) 
We denote with \x) ± a vector that is orthogonal to \x). 



Now we can define the concept of an orthogonal basis which will be 
very useful in finding the linear combination of vectors that give \x). 

Definition 9 An orthogonal basis of an N dimensional vector space 
V is a set of N linearly independent vectors such that each pair of 
vectors are orthogonal to each other. 

Example: In C the three vectors 


form an orthogonal basis. 

Planned end of 3 r lecture 

Now let us chose an orthogonal basis {\x) 1 , . . . , \x) N } of an N di- 
mensional vector space. For any arbitrary vector \x) G V we would like 
to find the coefficients Ai, . . . , Ajv such that 


£A^).Hz) • (1-38) 

Of course we can obtain the A, by trial and error, but we would like to 
find an efficient way to determine the coefficients Aj. To see this, let 
us consider the scalar product between \x) and one of the basis vectors 
\x)i. Because of the orthogonality of the basis vectors, we find 

(\x) i ,\x))=\ i (\x) i ,\x) i ) • (1.39) 

Note that this result holds true only because we have used an orthogonal 
basis. Using Eq. (1.39) in Eq. (1.38), we find that for an orthogonal 
basis any vector \x) can be represented as 

w =sft8K (L40) 

In Eq. (1.40) we have the denominator (\x) i , which makes the 
formula a little bit clumsy. This quantity is the square of what we 



Figure 1.7: A vector \x) = y^j in R 2 . From plane geometry we know 

that its length is \/ a 2 + b 2 which is just the square root of the scalar 
product of the vector with itself. 

usually call the length of a vector. This idea is illustrated in Fig. 1.7 
which shows a vector \x) = (jf) in the two-dimensional real vector 

space R 2 . Clearly its length is y/a 2 + b 2 . What other properties does 
the length in this intuitively clear picture has? If I multiply the vector 
by a number a then we have the vector a\x) which evidently has the 
length \Ja 2 a 2 + a 2 b 2 = \a\\/a 2 + b 2 . Finally we know that we have a 
triangular inequality. This means that given two vectors \x) 1 = (j^j 

and \x) 2 = (^2) the l en gth of the \x) 1 + \x) 2 is smaller than the sum 
of the lengths of \x) 1 and \x) 2 . This is illustrated in Fig. 1.8. In 
the following I formalize the concept of a length and we will arrive at 
the definition of the norm of a vector \x) v The concept of a norm is 
important if we want to define what we mean by two vectors being close 
to one another. In particular, norms are necessary for the definition of 
convergence in vector spaces, a concept that I will introduce in the 
next subsection. In the following I specify what properties a norm of a 
vector should satisfy. 

Definition 10 A norm on a vector space V associates with every \x) G 
V a real number with the properties. 

1. V|x> G V : > and \\\x)\\ = \x) = \0) . 

2. y\x) G V, a G C : ||o;|x)]| = |a| • |||x)||. (linearity) 



Figure 1.8: The triangular inequality illustrated for two vectors \x) and 
\y). The length of \x) + \y) is smaller than the some of the lengths of 
\x) and \y). 

3. y\x), \y) e V : \\\x) + \y)\\ < \\\x)\\ + \\\y)\\. (triangular 

A vector space with a norm defined on it is also called a normed 
vector space. The three properties in Definition 10 are those that 
you would intuitively expect to be satisfied for any decent measure of 
length. As expected norms and scalar products are closely related. In 
fact, there is a way of generating norms very easily when you already 
have a scalar product. 

Lemma 11 Given a scalar product on a complex vector space, we can 
define the norm of a vector \x) by 

\\\x)\\ = J(\x),\x)) . (1.41) 


• Properties 1 and 2 of the norm follow almost trivially from the 
four basic conditions of the scalar product. □ 



• The proof of the triangular inequality uses the Schwarz inequality. 

IH + |V>I| 2 = \(\x) + \y),\x) + \y))\ 

= \(\x),\x) + \y)) + (\y),\x) + \y))\ 

< \(\x),\x) + \y))\ + \(\y),\x) + \y))\ 

< llk>ll-llk> + |y>ll + llly>ll-llk> + |y>Mi^2) 

Dividing both sides by \\\x) + \y)\\ yields the inequality. This 
assumes that the sum \x) + \y) ^ \0). If we have \x) + \y) = \0) 
then the Schwarz inequality is trivially satisfied. □ 

Lemma 11 shows that any unitary vector space can canonically 
(this means that there is basically one natural choice) turned into a 
normed vector space. The converse is, however, not true. Not every 
norm gives automatically rise to a scalar product (examples will be 
given in the exercises). 

Using the concept of the norm we can now define an orthonormal 
basis for which Eq. (1.40) can then be simplified. 

Definition 12 An orthonormal basis of an N dimensional vector 
space is a set of N pairwise orthogonal linearly independent vectors 
{\x) 1 , . . . , \x) N } where each vector satisfies |||a;)j|| 2 = ja;)^) = 1, 

i.e. they are unit vectors. For an orthonormal basis and any vector \x) 
we have 

N N 

\ X ) = \x))\x)i = Y, a i\ X )i > ( L43 ) 

i=l i=l 

where the components of \x) with respect to the basis {\x) 1 , . . . , \x) N } 
are the = (\x)i, \x)). 

Remark: Note that in Definition 12 it was not really necessary to de- 
mand the linear independence of the vectors . . . , \x) N } because 
this follows from the fact that they are normalized and orthogonal. Try 
to prove this as an exercise. 

Now I have defined what an orthonormal basis is, but you still do 
not know how to construct it. There are quite a few different methods 



to do so. I will present the probably most well-known procedure which 
has the name Gram-Schmidt procedure. 

This will not be presented in the lecture. You should study this at home. 

There will be an exercise on this topic in the Rapid Feedback class. 

The starting point of the Gram-Schmidt orthogonalization proce- 
dure is a set of linearly independent vectors S = {\x) 1 , . . . , \x) n } . Now 
we would like to construct from them an orthonormal set of vectors 
{|e) 1 , . . . , |e) n }. The procedure goes as follows 

First step We chose \f) 1 = \x) 1 and then construct from it the nor- 
malized vector |e) 1 = 

Comment: We can normalize the vector \x) 1 because the set S is 
linearly independent and therefore \x) 1 ^ 0. 

Second step We now construct \f) 2 = \x) 2 — (\e) 1: \x) 2 )\e) l and from 
this the normalized vector |e) 2 = l/^/lll/^ll- 
Comment: 1) \f) 2 ^ \0) because \x) 1 and \x) 2 are linearly inde- 

2) By taking the scalar product (|e) 2 , |e) 1 ) we find straight away 
that the two vectors are orthogonal. 

k-th step We construct the vector 


Because of linear independence of S we have \f) k ^ \0). The 
normalized vector is then given by 



' ' k \\\f) k \\ ' 

It is easy to check that the vector \e) k is orthogonal to all \e) i 
with i < k. 

n-th step With this step the procedure finishes. We end up with a 
set of vectors {|e) 1 , . . . , |e) n } that are pairwise orthogonal and 




1.2.5 Completeness and Hilbert spaces 

In the preceding sections we have encountered a number of basic ideas 
about vector spaces. We have introduced scalar products, norms, bases 
and the idea of dimension. In order to be able to define a Hilbert space, 
the state space of quantum mechanics, we require one other concept, 
that of completeness, which I will introduce in this section. 

What do we mean by complete? To see this, let us consider se- 
quences of elements of a vector space (or in fact any set, but we are 
only interested in vector spaces). I will write sequences in two different 

{\x)i}i= ,...,oo = (\x) , \x) 1 ,\x) 2 ...) ■ (1.44) 

To define what we mean by a convergent sequences, we use norms 
because we need to be able to specify when two vectors are close to 
each other. 

Definition 13 A sequence {|:r)i}j=o,...,oo of elements from a normed 
vector space V converges towards a vector \x) e V if for all e > 
there is an n such that for all n > n we have 

\\\x) - \x) n \\ < e . (1.45) 

But sometimes you do not know the limiting element, so you would 
like to find some other criterion for convergence without referring to 
the limiting element. This idea led to the following 

Definition 14 A sequence {|x)i}i=o,...,oo of elements from a normed 
vector space V is called a Cauchy sequence if for all e > there is 
an no such that for all m, n > no we have 

\\\x) m - \x) n \\ < e . (1.46) 

Planned end of 4 th lecture 

Now you can wonder whether every Cauchy sequence converges. 
Well, it sort of does. But unfortunately sometimes the limiting ele- 
ment does not lie in the set from which you draw the elements of your 



sequence. How can that be? To illustrate this I will present a vector 
space that is not complete! Consider the set 

V = {\x) : only finitely many components of \x) are non-zero} . 

An example for an element of V is \x) = (1, 2, 3, 4, 5, 0, . . .). It is now 
quite easy to check that V is a vector-space when you define addition 
of two vectors via 

\x) + \y) = (x 1 + y 1 ,x 2 + y 2 ,.. .) 

and the multiplication by a scalar via 

c\x) = (cx 1 , cx 2 , . . .) . 

Now I define a scalar product from which I will then obtain a norm via 
the construction of Lemma 11. We define the scalar product as 


(\ x ), \y)) = Y. x lVk ■ 


Now let us consider the series of vectors 

k>i = (i,o,o,o,...) 

\x) 2 = (1,^,0,...) 
/ 1 1 

F> 3 = (i^'i' '---) 

\ x h li '2'4'8'"' J 

\ x )k 

1 1 

(1; 9 ) • • • ) 9 fc_i ' 0, . . .) 

2' 7 2 

For any no we find that for m > n > no we have 

1 1 1 

\ x )m~ \ x )n\\ = 11(0, •••,0, 2^'"'' 2^1'°'"' 



Therefore it is clear that the sequence {\x) k }k=i,...<x is a Cauchy se- 
quence. However, the limiting vector is not a vector from the vector 



space V, because the limiting vector contains infinitely many nonzero 

Considering this example let us define what we mean by a complete 
vector space. 

Definition 15 A vector space V is called complete if every Cauchy 
sequence of elements from the vector space V converges towards an 
element ofV. 

Now we come to the definition of Hilbert spaces. 

Definition 16 A vector space H is a Hilbert space if it satisfies the 
following two conditions 

1. H is a unitary vector space. 

2. H is complete. 

Following our discussions of the vectors spaces, we are now in the 
position to formulate the first postulate of quantum mechanics. 

Postulate 1 The state of a quantum system is described by a vector 
in a Hilbert space 7i. 

Why did we postulate that the quantum mechanical state space is 
a Hilbert space? Is there a reason for this choice? 

Let us argue physically. We know that we need to be able to rep- 
resent superpositions, i.e. we need to have a vector space. From the 
superposition principle we can see that there will be states that are not 
orthogonal to each other. That means that to some extent one quan- 
tum state can be 'present' in another non-orthogonal quantum state - 
they 'overlap'. The extent to which the states overlap can be quantified 
by the scalar product between two vectors. In the first section we have 
also seen, that the scalar product is useful to compute probabilities 
of measurement outcomes. You know already from your second year 



course that we need to normalize quantum states. This requires that 
we have a norm which can be derived from a scalar product. Because 
of the obvious usefulness of the scalar product, we require that the 
state space of quantum mechanics is a vector space equipped with a 
scalar product. The reason why we demand completeness, can be seen 
from a physical argument which could run as follows. Consider any 
sequence of physical states that is a Cauchy sequence. Quite obviously 
we would expect this sequence to converge to a physical state. It would 
be extremely strange if by means of such a sequence we could arrive 
at an unphysical state. Imagine for example that we change a state by 
smaller and smaller amounts and then suddenly we would arrive at an 
unphysical state. That makes no sense! Therefore it seems reasonable 
to demand that the physical state space is complete. 

What we have basically done is to distill the essential features of 
quantum mechanics and to find a mathematical object that represents 
these essential features without any reference to a special physical sys- 

In the next sections we will continue this programme to formulate 
more principles of quantum mechanics. 

1.2.6 Dirac notation 

In the following I will introduce a useful way of writing vectors. This 
notation, the Dirac notation, applies to any vector space and is very 
useful, in particular it makes life a lot easier in calculations. As most 
quantum mechanics books are written in this notation it is quite im- 
portant that you really learn how to use this way of writing vectors. If 
it appears a bit weird to you in the first place you should just practise 
its use until you feel confident with it. A good exercise, for example, is 
to rewrite in Dirac notation all the results that I have presented so far. 

So far we have always written a vector in the form \x). The scalar 
product between two vectors has then been written as (\x), \y)). Let us 
now make the following identification 

\x) <-> \x) . (1.47) 

We call \x) a ket. So far this is all fine and well. It is just a new 
notation for a vector. Now we would like to see how to rewrite the 



scalar product of two vectors. To understand this best, we need to talk 
a bit about linear functions of vectors. 

Definition 17 A function f : V — > C from a vector space into the 
complex numbers is called linear if for any \(f>) G V and any a, (5 G 
C we have 

/(<#) + m) = af(\1>)) + Pf(\m (1.48) 

With two linear function f\, f'2 also the linear combination fif\ + uf'2 
is a linear function. Therefore the linear functions themselves form a 
vector space and it is even possible to define a scalar product between 
linear functions. The space of the linear function on a vector space V 
is called the dual space V*. 

Now I would like to show you an example of a linear function which I 
define by using the scalar product between vectors. I define the function 
f\4>) '■ V — > C, where \<j>) £ V is & fixed vector so that for all G V 

f W m):=(\<P),m • (1.49) 

Now I would like to introduce a new notation for f^y From now on I 
will identify 

f\4) ~ (01 (1-50) 
and use this to rewrite the scalar product between two vectors \<p), \ip) 

(^>:=^|(|^» = (|^>,|^» . (1.51) 

The object (<p\ is called bra and the Dirac notation is therefore some- 
times called braket notation. Note that while the ket is a vector in 
the vector space V, the bra is an element of the dual space V*. 

At the moment you will not really be able to see the usefulness 
of this notation. But in the next section when I will introduce linear 
operators, you will realize that the Dirac notation makes quite a few 
notations and calculations a lot easier. 

1.3 Linear Operators 

So far we have only dealt with the elements (vectors) of vector spaces. 
Now we need to learn how to transform these vectors, that means how 



to transform one set of vectors into a different set. Again as quantum 
mechanics is a linear theory we will concentrate on the description of 
linear operators. 

1.3.1 Definition in Dirac notation 

Definition 18 An linear operator A : Ti — > Ti associates to every 
vector \ip) &Ti a vector A\ip) e Ti such that 

i(A|^)+/i|0)) = Ai|^)+M|0) (1.52) 

for all |0) G Ti and X, /i e C. 

Planned end of 5 th lecture 

A linear operator A : Ti — > 7i can be specified completely by de- 
scribing its action on a basis set of 7Y. To see this let us chose an 
orthonormal basis {|ej)|i = 1, . . . ,N}. Then we can calculate the ac- 
tion of A on this basis. We find that the basis {|ej)|i = 1, . . . , A^} is 
mapped into a new set of vectors = 1, . . . , A^} following 

\fi) := A\<*) . (1.53) 

Of course every vector can be represented as a linear combination 
of the basis vectors {\ei)\i = 1, . . . , A^}, i.e. 

\h) = Y, A ^k) ■ (1.54) 


Combining Eqs. (1.53) and (1.54) and taking the scalar product with 
\ej) we find 

A H = (ej\J2( A kM) (1-55) 


= (ej\fi) 

= (ej\A\ei) . (1.56) 

The Aji are called the matrix elements of the linear operator A with 
respect to the orthonormal basis {|ej)|i = 1, . . . ,N}. 



I will now go ahead and express linear operators in the Dirac nota- 
tion. First I will present a particularly simple operator, namely the unit 
operator 1, which maps every vector into itself. Surprisingly enough 
this operator, expressed in the Dirac notation will prove to be very 
useful in calculations. To find its representation in the Dirac notation, 
we consider an arbitrary vector |/) and express it in an orthonormal 
basis {\ei)\i = 1,...,N}. We find 

N N 

l/> = E/;|e;> = Ele;>tel/> ■ ( L57 ) 

3=1 3=1 

To check that this is correct you just form the scalar product between 
|/) and any of the basis vectors |e^) . Now let us rewrite Eq. (1.57) a 
little bit, thereby defining the Dirac notation of operators. 

N N 

l/) = EM( e ,l/)=:(EM(e J |)|/) (1-58) 

3=1 3=1 

Note that the right hand side is defined in terms of the left hand side. 
The object in the brackets is quite obviously the identity operator be- 
cause it maps any vector |/) into the same vector |/). Therefore it is 
totally justified to say that 


This was quite easy. We just moved some brackets around and we 
found a way to represent the unit operator using the Dirac notation. 
Now you can already guess how the general operator will look like, but 
I will carefully derive it using the identity operator. Clearly we have 
the following identity 

A = 1A1 . (1.60) 

Now let us use the Dirac notation of the identity operator in Eq. (1.59) 
and insert it into Eq. (1.60). We then find 

N N 

A = (EI^IM(Eb><e*l) 

3=1 k=l 




= Y,\ e j)(( e M\ek))(e k \ 


= T,(( e M\ e k))\ e j)( e k\ 


= Y, A ik\dj){e k \ . 



Therefore you can express any linear operator in the Dirac notation, 
once you know its matrix elements in an orthonormal basis. 

Matrix elements are quite useful when we want to write down lin- 
ear operator in matrix form. Given an orthonormal basis {|ej)|i = 
1, . . . , N} we can write every vector as a column of numbers 

( 91 \ 

l<?> = E^h> = ; • (i.62) 

Then we can write our linear operator in the same basis as a matrix 

f A u ... A 1N \ 

V A N i 


INN / 

To convince you that the two notation that I have introduced give the 
same results in calculations, apply the operator A to any of the basis 
vectors and then repeat the same operation in the matrix formulation. 

1.3.2 Adjoint and Hermitean Operators 

Operators that appear in quantum mechanics are linear. But not all 
linear operators correspond to quantities that can be measured. Only 
a special subclass of operators describe physical observables. In this 
subsection I will describe these operators. In the following subsection I 
will then discuss some of their properties which then explain why these 
operators describe measurable quantities. 

In the previous section we have considered the Dirac notation and 
in particular we have seen how to write the scalar product and matrix 



elements of an operator in this notation. Let us reconsider the matrix 
elements of an operator A in an ortho normal basis {|ej)|i = 1, . . . , N}. 
We have 

(e i |(i| ej -» = «e i |i)| ej -> (1.64) 

where we have written the scalar product in two ways. While the left 
hand side is clear there is now the question, what the bra (ei\A on the 
right hand side means, or better, to which ket it corresponds to. To see 
this we need to make the 

Definition 19 The adjoint operator A^ corresponding to the linear 
operator A is the operator such that for all \x),\y) we have 

(ti\x),\y)):=(\x),A\y)) , (1.65) 

or using the complex conjugation in Eq. (1.65) we have 

{y\ti\x) := (x\A\y)* . (1.66) 

In matrix notation, we obtain the matrix representing A^ by transposi- 
tion and complex conjugation of the matrix representing A. (Convince 
yourself of this). 

A- (J?) (1*7, 


A1 = ( -2i 2 ) t 1 - 68 ) 
The following property of adjoint operators is often used. 

Lemma 20 For operators A and B we find 

(AB)* = . (1.69) 

Proof: Eq. (1.69) is proven by 

(\x),(ABy\y) = ((AB)\x),\y)) 

= (A(B\x)), \y)) Now use Def. 19. 

= (B\x)),A j \y)) Use Def. 19 again. 

= (\x)),&A*\y)) 



As this is true for any two vectors \x) and \y) the two operators (AB)t 
and fiU* are equal □. 

It is quite obvious (see the above example) that in general an op- 
erator A and its adjoint operator A* are different. However, there are 
exceptions and these exceptions are very important. 

Definition 21 An operator A is called Hermitean or self-adjoint if 

it is equal to its adjoint operator, i.e. if for all states \x), \y) we have 

(y\A\x) = (x\A\y)* . (1.70) 

In the finite dimensional self-adjoint operator is the same as 

a Hermitean operator. In the infinite-dimensional case Hermitean and 
self-adjoint are not equivalent. 

The difference between self- adjoint and Hermitean is related to the 
domain of definition of the operators A and A' which need not be the 
same in the infinite dimensional case. In the following I will basically 
always deal with finite-dimensional systems and I will therefore usually 
use the term Hermitean for an operator that is self-adjoint. 

1.3.3 Eigenvectors, Eigenvalues and the Spectral 

Hermitean operators have a lot of nice properties and I will explore some 
of these properties in the following sections. Mostly these properties 
are concerned with the eigenvalues and eigenvectors of the operators. 
We start with the definition of eigenvalues and eigenvectors of a linear 

Definition 22 A linear operator A on an N -dimensional Hilbert space 
is said to have an eigenvector |A) with corresponding eigenvalue A if 

A\X) = X\X) , (1.71) 

or equivalently 

(A -A!) | A) = 




This definition of eigenvectors and eigenvalues immediately shows 
us how to determine the eigenvectors of an operator. Because Eq. 
(1.72) implies that the N columns of the operator A — AIL are linearly- 
dependent we need to have that 

det(A - AIL)) = . (1.73) 

This immediately gives us a complex polynomial of degree N. As we 
know from analysis, every polynomial of degree N has exactly N solu- 
tions if one includes the multiplicities of the eigenvalues in the counting. 
In general eigenvalues and eigenvectors do not have many restriction for 
an arbitrary A. However, for Hermitean and unitary (will be defined 
soon) operators there are a number of nice results concerning the eigen- 
values and eigenvectors. (If you do not feel familiar with eigenvectors 
and eigenvalues anymore, then a good book is for example: K.F. Riley, 
M.P. Robinson, and S.J. Bence, Mathematical Methods for Physics and 

We begin with an analysis of Hermitean operators. We find 

Lemma 23 For any Hermitean operator A we have 

1. All eigenvalues of A are real. 

2. Eigenvectors to different eigenvalues are orthogonal. 

Proof: 1.) Given an eigenvalue A and the corresponding eigenvector 
| A) of a Hermitean operator A. Then we have using the hermiticity of 

A* = (A|i|A)* = (Ali^A) = (A|i|A) = A (1.74) 

which directly implies that A is real □ . 

2.) Given two eigenvectors |A), for different eigenvalues A and 
/i. Then we have 

A<A|/*> = (A<a*|A»* = mA\X)Y = {\\A\n) = h(\\im) (1.75) 

As A and \i are different this implies (A|/i) = 0. This finishes the proof □. 



Lemma 23 allows us to formulate the second postulate of quantum 
mechanics. To motivate the second postulate a little bit imagine that 
you try to determine the position of a particle. You would put down 
a coordinate system and specify the position of the particle as a set of 
real numbers. This is just one example and in fact in any experiment 
in which we measure a quantum mechanical system, we will always ob- 
tain a real number as a measurement result. Corresponding to each 
outcome we have a state of the system (the particle sitting in a partic- 
ular position), and therefore we have a set of real numbers specifying 
all possible measurement outcomes and a corresponding set of states. 
While the representation of the states may depend on the chosen basis, 
the physical position doesn't. Therefore we are looking for an object, 
that gives real numbers independent of the chosen basis. Well, the 
eigenvalues of a matrix are independent of the chosen basis. Therefore 
if we want to describe a physical observable its a good guess to use an 
operator and identify the eigenvalues with the measurement outcomes. 
Of course we need to demand that the operator has only real eigenval- 
ues. Thus is guaranteed only by Hermitean operators. Therefore we 
are led to the following postulate. 

Postulate 2 Observable quantum mechanical quantities are described 
by Hermitean operators A on the Hilbert space H. The eigenvalues aj 
of the Hermitean operator are the possible measurement results. 

-Planned end of 6* lecture 

Now I would like to formulate the spectral theorem for finite di- 
mensional Hermitean operators. Any Hermitean operator on an N- 
dimensional Hilbert space is represented by an N x N matrix. The 
characteristic polynomial Eq. (1.73) of such a matrix is of N-th order 
and therefore possesses N eigenvalues, denoted by A« and corresponding 
eigenvectors |Aj). For the moment, let us assume that the eigenvalues 
are all different. Then we know from Lemma 23 that the correspond- 
ing eigenvectors are orthogonal. Therefore we have a set of N pairwise 



orthogonal vectors in an iV-dimensional Hilbert space. Therefore these 
vectors form an orthonormal basis. From this we can finally conclude 
the following important 

Completeness theorem: For any Hermitean operator i on a 
Hilbert space Ti the set of all eigenvectors form an orthonormal basis 
of the Hilbert space H, i.e. given the eigenvalues A« and the eigenvectors 
|Aj) we find 

A = X)Ai|A i >(A i | (1.76) 


and for any vector \x) G Ti. we find coefficients Xi such that 

\x) = Y t x i \\) . (1.77) 


Now let us briefly consider the case for degenerate eigenvalues. This 
is the case, when the characteristic polynomial has multiple zero's. In 
other words, an eigenvalue is degenerate if there is more than one 
eigenvector corresponding to it. An eigenvalue A is said to be d- 
fold degenerate if there is a set of d linearly independent eigenvec- 
tors {|Ai), . . . , | Ad)} all having the same eigenvalue A. Quite obviously 
the space of all linear combinations of the vectors {|Ai), . . . , \ \d)} is 
a d- dimensional vector space. Therefore we can find an orthonormal 
basis of this vector space. This implies that any Hermitean operator 
has eigenvalues Ai, . . . , A& with degeneracies d(\i), . . . , d(Afc). To each 
eigenvectorAj I can find an orthonormal basis of d(Xi) vectors. There- 
fore the above completeness theorem remains true also for Hermitean 
operators with degenerate eigenvalues. 

Now you might wonder whether every linear operator A on an N 
dimensional Hilbert space has N linearly independent eigenvectors? It 
turns out that this is not true. An example is the 2x2 matrix 

which has only one eigenvalue A = 0. Therefore any eigenvector to this 
eigenvalue has to satisfy 



which implies that 6 = 0. But then the only normalized eigenvector is 
and therefore the set of eigenvectors do not form an orthonormal 

We have seen that any Hermitean operator A can be expanded in its 
eigenvectors and eigenvalues. The procedure of finding this particular 
representation of an operator is called diagonalization. Often we do 
not want to work in the basis of the eigenvectors but for example in 
the canonical basis of the vectors 

ei = 

/ 1 \ 

V o J 

( o \ 


V o J 

, e N = 

( \ 

V i J 


If we want to rewrite the operator A in that basis we need to find 
a map between the canonical basis and the orthogonal basis of the 

eigenvectors of A. If we write the eigenvectors |Aj) = X^=i a ji\ e j) then 
this map is given by the unitary operator (we will define unitary 
operators shortly) 





a 1N 




which obviously maps a vector |e») into the eigenvector corresponding 
to the eigenvalue |Aj). Using this operator U we find 


The operator in Eq. (1.79) maps orthogonal vectors into orthogonal 
vectors. In fact, it preserves the scalar product between any two vectors. 
Let us use this as the defining property of a unitary transformation. 



Now as promised the accurate definition of a unitary operator. 
Avoiding the subtleties of infinite dimensional spaces (which are again 
problems of the domain of definition of an operator) for the moment 
we have the following 

Definition 24 A linear operator U on a Hilbert space TL is called uni- 
tary if it is defined for all vectors \x), \y) in TL, maps the whole Hilbert 
space into the whole Hilbert space and satisfies 

(x\m\y) = (x\y) . (1.81) 

In fact we can replace the last condition by demanding that the 
operator satisfies 

ITU = 1 and Uin = t. (1.82) 

Eq. (1.81) implies that a unitary operator preserves the scalar prod- 
uct and therefore in particular the norm of vectors as well as the angle 
between any two vectors. 

Now let us briefly investigate the properties of the eigenvalues and 
eigenvectors of unitary operators. The eigenvalues of a unitary oper- 
ators are in general not real, but they not completely arbitrary. We 

Theorem 25 Any unitary operator U on an N -dimensional Hilbert 
space TL has a complete basis of eigenvectors and all the eigenvalues are 
of the form e 1 ^ with real <p. 

Proof: I will not give a proof that the eigenvectors of U form a basis 
in TL. For this you should have a look at a textbook. What I will proof 
is that the eigenvalues of TL are of the form with real 0. To see this, 
we use Eq. (1.82). Be |A) an eigenvector of U to the eigenvalue A, then 

AC/t| A ) = f)tf)|A) = |A) . (1.83) 

This implies that A ^ because otherwise the right-hand side would 
be the null- vector, which is never an eigenvector. From Eq. (1.83) we 

I = (X\U^\X) = (\\U\X)* = A* . (1.84) 



This results in 

|A| 2 = l^A = e i * . (1.85) 

1.3.4 Functions of Operators 

In the previous sections I have discussed linear operators and special 
subclasses of these operators such as Hermitean and unitary operators. 
When we write down the Hamilton operator of a quantum mechanical 
system we will often encounter functions of operators, such as the one- 
dimensional potential V(x) in which a particle is moving. Therefore it 
is important to know the definition and some properties of functions of 
operators. There are two ways of defining functions on an operator, one 
works particularly well for operators with a complete set of eigenvectors 
(Definition 26), while the other one works bests for functions that can 
be expanded into power series (Definition 27). 

Definition 26 Given an operator A with eigenvalues a; t and a complete 
set of eigenvectors |a») . Further have a function f : C — > C that maps 
complex numbers into complex numbers then we define 


f(A) :=J2f( a i)\ a i)( a i\ (1-86) 

Definition 27 Given a function f : C — > C that can be expanded into 
a power series 


= (1.87) 


then we define 


f(A) = J2M t . (1.88) 


Definition 28 The derivative of an operator function f(A) is defined 
viag(z) = f z (z) as 

^P=,(i). (1.89) 



Let us see whether the two definitions Def. 26 and 27 coincide for 
operators with complete set of eigenvectors and functions that can be 
expanded into a power series given in Eq. (1.87). 


f(A) = J2hA k 

oo N 

= E/fc(E%K)(%l) fe 
k=i j=i 

oo N 

k=l j=l 

N oo 

= E(EM*)K>(%I 

3=1 k=l 

For operators that do not have a complete orthonormal basis of eigen- 
vectors of eigenvectors it is not possible to use Definition 26 and we 
have to go back to Definition 27. In practise this is not really a prob- 
lem in quantum mechanics because we will always encounter operators 
that have a complete set of eigenvectors. 

As an example consider a Hermitean operator A with eigenvalues 
dk and eigenvectors \a k ) and compute U = e lA . We find 


U = e iA = Y / e^\a k )(a k \ . (1.91) 

This is an operator which has eigenvalues of the form e iak with real a k . 
Therefore it is a unitary operator, which you can also check directly 
from the requirement UW = 1 = WU. In fact it is possible to show 
that every unitary operator can be written in the form Eq. (1.91). This 
is captured in the 

Lemma 29 To any unitary operator U there is a Hermitean operator 
H such that 

U = e ifl . (1.92) 




1) Show that for any unitary operator U we have f(WAU) = Wf(A)t) 
Proof: We use the fact that UU* = 1 to find 


= Y.hU ] A k U 



= rf(£f k A k )U 


= U j f(A)U . 

If you have functions, then you will also expect to encounter deriva- 
tives of functions. Therefore we have to consider how to take derivatives 
of matrices. To take a derivative we need to have not only one operator, 
but a family of operators that is parametrized by a real parameter s. 
An example is the set of operators of the form 

^)=(1+: ;: s s ) ■ 

Another example which is familiar to you is the time evolution operator 


Now we can define the derivative with respect to s in complete 
analogy to the derivative of a scalar function by 

dA A(s + As)-A(s) 

— Is) := hm — - 1 — . (1.94) 

ds As^o As v ' 

This means that we have defined the derivative of an operator compo- 
nent wise. 

Now let us explore the properties of the derivative of operators. 
First let us see what the derivative of the product of two operators is. 
We find 

Property: For any two linear operators A(s) and B(s) we have 
d(AB) . . dA j, ,dB . . 



This looks quite a lot like the product rule for ordinary functions, except 
that now the order of the operators plays a crucial role. 

Planned end of 7 th lecture 

You can also have functions of operators that depend on more than 
one variables. A very important example is the commutator of two 

Definition 30 For two operators A and B the commutator [A, B] is 
defined as 

[A, B] = AB-BA . (1.96) 

While the commutator between numbers (lxl matrices) is always 
zero, this is not the case for general matrices. In fact, you know al- 
ready a number of examples from your second year lecture in quantum 
mechanics. For example the operators corresponding to momentum 
and position do not commute, i.e. their commutator is nonzero. Other 
examples are the Pauli spin-operators 

ao = (o l) ai= (l o) 

_°0 (-) 

For i,j = 1, 2, 3 they have the commutation relations 

[<7;,rx,] = ie ijk a k . (1.98) 

where is the completely antisymmetric tensor. It is defined by ei 2 3 = 
1 and changes sign, when two indices are interchanged, for example 

There are some commutator relations that are quite useful to know. 

Lemma 31 For arbitrary linear operators A, B, C on the same Hilbert 
space we have 

[AB,C] = A[B,C] + [A,C]B (1.99) 
= [A,[B,C}} + [B,[C,A}} + [C,[A,B]} (1.100) 



Proof: By direct inspection of the equations □. 

Commuting observables have many useful and important proper- 
ties. Of particular significance for quantum physics is the following 
Lemma 32 because it guarantees that two commuting observables can 
be simultaneously measured with no uncertainty. 

Lemma 32 Two commuting observables A and B have the same eigen- 
vectors, i.e. they can be diagonalized simultaneously. 

Proof: For simplicity we assume that both observables have only non- 
degenerate eigenvalues. Now chose a basis of eigenvectors (|a;)} that 
diagonalizes A. Now try to see whether the |aj) are also eigenvectors 
of B. Using [A, B\ we have 

A{B\a i )) = BA\a l )=a l {B\a l )) . (1.101) 

This implies that B\ai) is an eigenvector of A with eigenvalue a«. As 
the eigenvalue is non-degenerate we must have 

Bla^bildi) (1.102) 

for some hi. Therefore |aj) is an eigenvector to B □. 

There is a very nice relation between the commutator and the 
derivative of a function. First I define the derivative of an operator 
function with respect to the operator. 

Lemma 33 Given two linear operators A and B which have the com- 
mutator [B,A] = 1L. Then for the derivative of an operator function 
f(A) we find 

[B,f(A)] = i(A) . (1.103) 

Proof: Remember that a function of an operator is defined via its 
expansion into a power series, see Eq. (1.88). Therefore we find 

[B,f(A)\ = [B,J2f k A k ] 





Now we need to evaluate the expression We proof by induc- 

tion that [-B,A n ] = nA n ~ x . For n — 1 this is true. Assume that the 
assumption is true for n. Now start with n + 1 and reduce it to the 
case for n. Using Eq. (1.99) we find 

[B, i (n+1) ] = [B, A n A] = [B, A n ]A + A n [B, A] . (1.104) 

Now using the [B, A] = 1 and the induction assumption we find 

[B, A in+1) ] = nk l - l A + A n = (n + l)A n . (1.105) 

Now we can conclude 

[BJ(A)\ = J2f k [B,A k } 



dA K 

This finishes proof. □ 

A very useful property is 

Lemma 34 (Baker-Campbell-Haussdorff) For general operators A and 
B we have 

e 6 Ae- 6 = A+[B,A\ + \[B,[B,A\\ + ... . (1.106) 

For operators such that [B, [B, A]] = we have the simpler version 

e^Ae-v = A+ [B,A] . (1.107) 

Proof: Define the function of one real parameter a 

f(a) = e a *Ae- at} . (1.108) 

We can expand this function around a = into a Taylor series f{a) = 
J2^=o j^{ a )\a=Q and therefore we need to determine the derivatives 



of the function f{a). We find 



The rest of the proof follows by induction. The proof of Eq. (1.107) 
follows directly from Eq. (1.106). n 

1.4 Operators with continuous spectrum 

In the preceding sections I have explained the basic ideas of Hilbert 
spaces and linear operators. All the objects I have been dealing with so 
far have been finite dimensional an assumption that greatly simplified 
the analysis and made the new ideas more transparent. However, in 
quantum mechanics many systems are actually described by an infinite 
dimensional state space and the corresponding linear operators are infi- 
nite dimensional too. Most results from the preceding section hold true 
also for the infinite dimensional case, so that we do not need to learn 
too many new things. Nevertheless, there are properties that require 
some discussion. 

1.4.1 The position operator 

The most natural place where an infinite dimensional state space ap- 
pears is in the analysis of a particle in free space. Therefore let us 
briefly reconsider some aspects of wave mechanics as you have learnt 
them in the second year course. The state of a particle in free space 
(maybe moving in a potential) is described by the square-integrable 
wave-function ip(x). The question is now as to how we connect the 
wave-function notation with the Dirac notation which we have used to 
develop our theory of Hilbert spaces and linear operators. 

Let us remember what the Dirac notation for finite dimensional 
systems means mathematically. Given the ket- vector |0) for the state 



of a finite dimensional system, we can find the components of this ket- 
vector with respect to a certain basis {|ej)}. The i-th component is 
given by the complex number (ei\<f>). Therefore in a particular basis it 
makes sense to write the state \<p) as a column vector 


V ( e n\<t>) J 


Let us try to transfer this idea to the wave-function of a particle in 
free space. What we will do is to interpret ip{x) as the component of a 
vector with infinitely many components. Informally written this means 






where we have given the column vector a name, Obviously the 

set of vectors defined in this way form a vector space as you can easily 
check. Of course we would like to have a scalar product on this vector 
space. This is introduced in complete analogy to the finite dimensional 
case. There we had 

(|0>,|^>)H0l^> = E<<^><e^> • (1.111) 


We just replace the summation over products of components of the two 
vectors by an integration. We have (see also Eq (1.29) 

dx^ix^x) . (1.112) 

Now we have a space of vectors (or square integrable wave functions) 
TL equipped with a scalar product. Indeed it turns out to be a complete 
space (without proof) so that we have a Hilbert space. 

Now that we have introduced ket vectors we also need to define 
bra-vectors. In the finite dimensional case we obtained the bra vector 
via the idea of linear functionals on the space of state vectors. Let us 


repeat this procedure now for the the case of infinite dimensions. We 
define a linear functional (cf>\ by 

M(M) = <^) = (|0,M) . (1.113) 

Now 1 would like to investigate a particular ket vector (linear func- 
tional) that will allow as to define a position state. We define a linear 
functional {x \ by 

(x \^)) = (x \^):=^x ) . (1.114) 

We are already writing this functional very suggestively as a bra-vector. 
Thats perfectly ok, and we just have to check that the so defined func- 
tional is indeed linear. Of course we would like to interpret the left 
hand side of Eq. (1.114) as a scalar product between two ket's, i.e. 

(\x ),\if>)):={x \i/>)='>f>M • (1.115) 

What does the ket \xq) corresponding to the bra (xq\ mean? Which 
wave- function 6* (x) does it correspond to? Using the scalar product 
Eq. (1.112), we have 

dx5* xo (x)^(x) = (|x ), m = (x \1>) = iP(x ) (1.116) 

This means that the function <5* (x) has to act like a delta-function! 
The wave-function corresponding to the bra (x \ is a delta-function. A 
delta-function however, is not square-integrable! Therefore it cannot be 
an element of the Hilbert space of square integrable functions. However, 
as we have seen it would be quite convenient to use these wave-functions 
or states. Therefore we just add them to our Hilbert space, although we 
will often call them improper states or wave-functions. In fact we can 
use basically all the rules that we have learned about finite dimensional 
Hilbert-spaces also for these improper states. All we need to demand 
is the following rule for the scalar product 

(i;\xo):=((x o m*=r(xo) ■ (1.H7) 
Now I can write for arbitrary kets \<f>), e H 

(tp\ip) — J (j)*(x)ip(x)dx = J {(f>\x){x\ip)dx = (<f>\(J \x) (x\dx)\ip) . 




Then we can conclude 

J \x)(x\dx = 1 . (1.119) 

Inserting this identity operator in (x\ip), we obtain the orthogonality 
relation between position kets 

J 5{x - x')4>(x')dx' = il>(x) = (x\i/>) = J (x\x') (x'\^)dx' 

= J (x\x')ip{x')dx' . 

Therefore we have 

(x\x) = 5(x-x') . (1.120) 

Now we can derive the form of the position operator from our knowledge 
of the definition of the position expectation value 

(ip\x\ip) := J x\ip{x)\ 2 dx 

= J (ip\x)x{x\ip)dx 

= mjx\x)(dxm , (1.121) 

where we defined the position operator 

x = J x\x)(x\dx = x ] . (1.122) 

Now you see why the improper position kets are so useful. In this basis 
the position operator is automatically diagonal. The improper position 
kets \xq) are eigenvectors of the position operator 

x\xq) = xo\xo) ■ (1.123) 

This makes sense, as the position kets describe a particle that is per- 
fectly localized at position x . Therefore a position measurement should 
always give the result xq. So far we have dealt with one-dimensional 
systems. All of the above considerations can be generalized to the 
d-dimensional case by setting 

\x) = x x e x + . . . + x d e d . (1.124) 

The different components of the position operator commute are as- 
sumed to commute. Planned end of „** lccture 


1.4.2 The momentum operator 

Now we are going to introduce the momentum operator and momentum 
eigenstates using the ideas of linear functionals in a similar fashion to 
the way in which we introduced the position operator. Let us introduce 
the linear functional (p\ defined by 

(pM := - J= J e-*°'*il>(x)dx . (1.125) 

Now we define the corresponding ket by 

<p|V>* = ^ip) =: (Vb> • (1-126) 

Combining Eq. (1.125) with the identity operator as represented in Eq. 
(1.119) we find 

j e -W^ (x)rfx = {m = J {p \ x)W)dx (1.127) 

Therefore we find that the state vector \p) represents a plane wave, 

(x\p) = —^ e ipx / h . (1.128) 

As Eq. (1.128) represents a plane wave with momentum p it makes 
sense to call \p) a momentum state and expect that it is an eigenvector 
of momentum operator p to the eigenvalue p. Before we define the mo- 
mentum operator, let us find the decomposition of the identity operator 
using momentum eigenstates. To see this we need to remember from 
the theory of delta-functions that 

JL f e ip{x-y)/n dp = 5 ^ x _ y ^ ( 1129 ) 
znh J 

Then we have for arbitrary \x) and \y) 

(x\y) = 5{x-y) = ^ h J dpe^^ h = <*!(/ \p)(p\dp)\y) , (1-130) 
and therefore 

' \p)(p\dp= 1 . (1.131) 




The orthogonality relation between different momentum kets can be 
found by using Eq. (1.131) in Eq. (1.125). 

(p|^) = (p|!|V>) = J (p\p'){p'\ip)dp' (1.132) 

so that 

(p\p'} = 5(p - p') . (1.133) 

The momentum operator p is the operator that has as its eigenvec- 
tors the momentum eigenstates \p) with the corresponding eigenvalue 
p. This makes sense, because \p) describes a plane wave which has a 
perfectly defined momentum. Therefore we know the spectral decom- 
position which is 

P = Jp\p)(p\dp ■ (1.134) 

Clearly we have 

Pbo)=Pobo) • (1.135) 

Analogously to the position operator we can extend the momentum 
operator to the d-dimensional space by 

P = Piei + . . . +p d e d (1.136) 

The different components of the momentum operator are assumed to 

1.4.3 The position representation of the momen- 
tum operator and the commutator between 
position and momentum 

We have seen how to express the position operator in the basis of the 
improper position kets and the momentum operator in the basis of the 
improper momentum kets. Now I would like to see how the momentum 
operator looks like in the position basis. 

To see this, differentiate Eq. (1.128) with respect to x which gives 



Therefore we find 

{x\p\ip) = J (x\p)p(p\tp)dp 

= ~7T I (x\p)(p\^)dp 

= *{x\(f\p)<p\m 


i dx 

In position representation the momentum operator acts like the differ- 
ential operator, i.e. 

Knowing this we are now able to derive the commutation relation 
between momentum and position operator. We find 


(x\(xp — px\ip) 

7 x fa( x \^) - fa(. x ( x M) 


Therefore we have the Heisenberg commutation relations 

[x,p] = ihl . 



Chapter 2 

Quantum Measurements 

So far we have formulated two postulates of quantum mechanics. The 
second of these states that the measurement results of an observable 
are the eigenvalues of the corresponding Hermitean operator. However, 
we do not yet know how to determine the probability with which this 
measurement result is obtained, neither have we discussed the state of 
the system after the measurement. This is the object of this section. 

2.1 The projection postulate 

We have learned, that the possible outcomes in a measurement of a 
physical observable are the eigenvalues of the corresponding Hermitean 
operator. Now I would like to discuss what the state of the system 
after such a measurement result is. Let us guide by common sense. 
Certainly we would expect that if we repeat the measurement of the 
same observable, we should find the same result. This is in contrast 
to the situation before the first measurement. There we certainly did 
not expect that a particular measurement outcome would appear with 
certainty. Therefore we expect that the state of the system after the 
first measurement has changed as compared to the state before the 
first measurement. What is the most natural state that corresponds to 
the eigenvalue Oj of a Hermitean operator A? Clearly this is the cor- 
responding eigenvector in which the observable has the definite value 
<2j! Therefore it is quite natural to make the following postulate for 



observables with non-degenerate eigenvalues. 

Postulate 3 (a) The state of a quantum mechanical system after the 
measurement of observable A with the result being the non-degenerated 
eigenvalue a« is given by the corresponding eigenvector |a») . 

For observables with degenerate eigenvalues we have a problem. 
Which of the eigenvalues should we chose? Does it make sense to chose 
one particular eigenvector? To see what the right answer is we have to 
make clear that we are really only measuring observable A. We do not 
obtain any other information about the measured quantum mechanical 
system. Therefore it certainly does not make sense to prefer one over 
another eigenvector. In fact, if I would assume that we have to chose 
one of the eigenvectors at random, then I would effectively assume 
that some more information is available. Somehow we must be able to 
decide which eigenvector has to be chosen. Such information can only 
come from the measurement of another observable - a measurement 
we haven't actually performed. This is therefore not an option. On 
the other hand we could say that we chose one of the eigenvectors at 
random. Again this is not really an option, as it would amount to 
saying that someone chooses an eigenvector and but does not reveal to 
us which one he has chosen. It is quite important to realize, that a) not 
having some information and b) having information but then choosing 
to forget it are two quite different situations. 

All these problems imply that we have to have all the eigenvectors 
still present. If we want to solve this problem, we need to introduce a 
new type of operator - the projection operator. 

Definition 35 An operator P is called a projection operator if it 

1. P = Pt ; 

2. P = P 2 



Some examples for projection operators are 



If P is a projection operator, then also 1 — P is a projection 

3. P = i 

Exercise: Prove that the three examples above are projection opera- 

Lemma 36 The eigenvalues of a projection operator can only have the 
values or 1. 

Proof: For any eigenvector |A) of A we have 

From this we immediately obtain A = 0orA = l n . 

For a set of orthonormal vectors {|'0i)}i=i,...,iv a projection operator 
P = J2i=i projects a state onto the subspace spanned by 

the vectors {|'0i)}i=i,...,fe- 111 mathematics this statement is more clear. 
If we expand = X)£Li then 


A|A)=P|A) = P 2 |A) = A 2 |A) . 





This is exactly what we need for a more general formulation of the 
third postulate of quantum mechanics. We formulate the third postu- 
late again, but now for observables with degenerate eigenvalues. 



Postulate 3 (b) The state of a quantum mechanical system after the 
measurement of general observable A with the result being the possibly 
degenerated eigenvalue a,i is given by 

P#> • (2.3) 

where Pi is the projection operator on the subspace of H spanned by 
all the eigenvectors of A with eigenvalue ai, i.e. 

A = £I^>M • (2-4) 

Experiments have shown that this postulate describes the results 
of measurements extremely well and we therefore have to accept it at 
least as a good working assumption. I say this because there are quite a 
few people around who do not like this postulate very much. They ask 
the obvious question as to how and when the reduction of the quantum 
state is exactly taking place. In fact, we have not answered any of that 
in this discussion and I will not do so for a simple reason. Nobody 
really knows the answer to these questions and they are a subject of 
debate since a long time. People have come up with new interpretations 
of quantum mechanics (the most prominent one being the many-worlds 
theory) or even changes to the Schrodinger equations of quantum me- 
chanics themselves. But nobody has solved the problem satisfactorily. 
This is a bit worrying because this means that we do not understand 
an important aspect of quantum mechanics. However, there is a really 
good reason for using the projection postulate: It works superbly 
well when you want to calculate things. 

Therefore let us now continue with our exploration of the postulates 
of quantum mechanics. We now know what the measurement outcomes 
are and what the state of our system after the measurement is, but 
we still do not know what the probability for a specific measurement 
outcome is. To make any sensible guess about this, we need to consider 



the properties of probabilities to see which of these we would like to 
be satisfied in quantum mechanics. Once we know what are sensible 
requirements, then we can go ahead to make a new postulate. 

As you know, probabilities p{Aj) to obtain an outcome Ai (which 
is usually a set of possible results) are positive quantities that are not 
larger than unity, i.e. < p(Ai) < 1. In addition it is quite trivial to 
demand that the probability for an outcome corresponding to an empty 
set vanishes, i.e. p(0), while the probability for the set of all elements 
is unity, i.e. p(TL). 

These are almost trivial requirements. Really important is the be- 
haviour of probabilities for joint sets. What we definitively would like 
to have is that the probabilities for mutually exclusive events add up, 
i.e. if we are given disjoint sets A\ and A 2 and we form the union 
A 1 U A 2 of the two sets, then we would like to have 

p(A 1 UA 2 ) =p(A 1 )+p(A 2 ) . (2.5) 

In the following postulate I will present a definition that satisfies all of 
the properties mentioned above. 

Postulate 4 The probability of obtaining the eigenvalue aj in a mea- 
surement of the observable A is given by 

Pi = \\Pi\m 2 = ^\H^) ■ (2-6) 

For a non-degenerate eigenvalue with eigenvector this probabil- 
ity reduces to the well known expression 

1H = mW)\ 2 (2-7) 

It is easy to check that Postulate 4 indeed satisfies all the criteria 
that we demanded for a probability. The fascinating thing is, that it is 
essentially the only way of doing so. This very important theorem was 
first proved by Gleason in 1957. This came as quite a surprise. Only 
using the Hilbert space structure of quantum mechanics together with 



the reasonable properties that we demanded from the probabilities for 
quantum mechanical measurement outcomes we can't do anything else 
than using Eq. (2.6)! The proof for this theorem is too complicated 
to be presented here and I just wanted to justify the Postulate 4 a bit 
more by telling you that we cannot really postulate anything else. 

2.2 Expectation value and variance. 

Now that we know the measurements outcomes as well as the probabil- 
ities for their occurrence, we can use quantum mechanics to calculate 
the mean value of an observable as well as the variance about this mean 

What do we do experimentally to determine an mean value of an 
observable. First, of course we set up our apparatus such that it mea- 
sures exactly the observable A in question. Then we build a machine 
that creates a quantum system in a given quantum state over and over 
again. Ideally we would like to have infinitely many of such identically 
prepared quantum states (this will usually be called an ensemble). 
Now we perform our experiment. Our state preparer produces the first 
particle in a particular state and the particle enters our measure- 
ment apparatus. We obtain a particular measurement outcome, i.e. one 
of the eigenvalues Oj of the observable A. We repeat the experiment 
and we find a different eigenvalue. Every individual outcome of the 
experiment is completely random and after iV repetitions of the experi- 
ment, we will obtain the eigenvalue measurement outcome in Ni 
of the experiments, where X^iVj — N. After finishing our experiment, 
we can then determine the average value of the measurement outcomes, 
for which we find 

(*>*W = E f «i • (2-8) 


For large numbers of experiments, i.e. large N, we know that the ratio 
^ approaches the probability pi for the outcome a«. This probability 
can be calculated from Postulate 4 and we find 


(i4>^ = ^2^ a * = Eft a * = E a *^W> = ^W> • ( 2 - 9 ) 

i i i 



You should realize that the left hand side of Eq. (2.9) is conceptually 
quite a different thing from the right hand side. The left hand side 
is an experimentally determined quantity, while the right hand side 
is a quantum mechanical prediction. This is a testable prediction of 
Postulate 3 and has been verified in countless experiments. Now we 
are also able to derive the quantum mechanical expression for the vari- 
ance of the measurement results around the mean value. Given the 
probabilities pi and the eigenvalues Oj of the observable A we find 

(AA) 2 := - (A)) 2 = Y>a 2 - (^fta,) 2 = (A 2 ) - (A) 2 

After these definitions let us now analyse a very important property 
of measurement results in quantum mechanics. 

In classical mechanics we are able to measure any physical quantity to 
an arbitrary precision. In fact, this is also true for quantum mechanics! 
Nothing prevents us from preparing a quantum system in a very well 
defined position state and subsequently make a measurement that will 
tell us with exceeding precision where the particle is. The real difference 
to classical mechanics arise when we try to determine the values of two 
different observables simultaneously. Again, in classical mechanics we 
have no law that prevents us from measuring any two quantities with 
arbitrary precision. Quantum mechanics, however, is different. Only 
very special pairs of observables can be measured simultaneously to 
an arbitrary precision. Such observables are called compatible or com- 
muting observables. In general the uncertainties in the measurements 
of two arbitrary observables will obey a relation which makes sure that 
their product has a lower bound which is in general is unequal to zero. 
This relation is called the uncertainty relation. 

Theorem 37 For any two observables A and B we find for their un- 
certainties AX = \J (X 2 ) — (A) 2 the uncertainty relation 


2.3 Uncertainty Relations 







Proof: Let us define two ket's 

\<Pa) = {A-{A))\^) (2.12) 

\4b) = (B - (B)) |V> ■ (2.13) 
Using these vectors we find that 

^\(<Pa\<Pa)\-\(<Pb\<Pb)\=^AAB . (2.14) 

Now we can use the Schwarz inequality to find a lower bound on the 
product of the two uncertainties. We find that 

< V / I<<M<MH(<H<MI = AAAB . (2.15) 
For the real and imaginary parts of |(0a|0b)| we find 

M<t>A\<t> B ) = \({AB + BA)-2{A){B)) (2.16) 
Im{<f> A \<f> B ) = ^{[A,B]). (2.17) 
We can use Eqs. (2.16-2.17) in Eq. (2.15) to find 

AAAB > y/(Re(<i> A \<j> B ))* + (Im(<j> A \<j> B ))* > \\{[A,B})\ (2.18) 

This finishes the proof □ . 

From Eq. (2.11) we can see that a necessary condition for two 
observables to be measurable simultaneously with no uncertainty is 
that the two observables commute. In fact this condition is also a 
sufficient one. We have 

Theorem 38 Two observables A and B can be measured precisely, i.e. 
A A = AB = 0, exactly if they commute. 

Proof: From Eq. (2.11) we immediately see that AA = AB = im- 
plies that [A, B] = 0. For the other direction of the statement we need 
to remember that two commuting observables have the same eigen- 
vectors. If we assume our quantum mechanical system is in one of 



these eigenstates then we see that in Eqs. (2.12-2.13) that \<Pa) 
and \4>b) are proportional to \ip) and therefore proportional to each 
other. Then we only need to remember that in the case of proportional 
\4>a) and \4>b) we have equality in the Schwarz inequality which implies 
AA = AB=\\{[A,B])\=Q n . 

Now let us make clear what the uncertainty relation means, as there 
is quite some confusion about that. Imagine that we are having an en- 
semble of infinitely many identically prepared quantum systems, each 
of which is in state \ip). We would like to measure two observables A 
and B. Then we split the ensemble into two halves. On the first half 
we measure exclusively the observable A while on the second half we 
measure the observable B. The measurement of the two observables 
will lead to average values (A) and (B) which have uncertainties AA 
and AB. From this consideration it should be clear, that the uncer- 
tainties in the measurements of the two observables are not caused by 
a perturbation of the quantum mechanical system by the measurement 
as we are measuring the two observables on different systems which are 
in the same state. These uncertainties are an intrinsic property of 
any quantum state. Of course perturbations due to the measure- 
ment itself may increase the uncertainty but it is not the reason for 
the existence of the uncertainty relations. 

In that context I would also like to point out, that you will some- 
times find the uncertainty relation stated in the form 

AAAB > \([A,B})\ , (2.19) 

that is, where the right-hand side is twice as large as in Eq. (2.11). 
The reason for this extra factor of 2 is that the uncertainty relation Eq. 
(2.19) describes a different situation than the one stated in Theorem 37. 
Eq. (2.19) really applies to the simultaneous measurement of two non- 
commuting observables on one quantum system. This means that we 
perform a measurement of the observable A on every system of the en- 
semble and subsequently we perform a measurement of the observable 
B on every system of the ensemble. The measurement of the observable 
A will disturb the system and a subsequent measurement of B could 
therefore be more uncertain. Imagine, for example, that you want to 



determine the position of an electron and simultaneously the momen- 
tum. We could do the position measurement by scattering light from 
the electron. If we want to determine the position of the electron very 
precisely, then we need to use light of very short wavelength. However, 
photons with a short wavelength carry a large momentum. When they 
collide with the electron, then the momentum of the electron may be 
changed. Therefore the uncertainty of the momentum of the electron 
will be larger. This is an example for the problems that arise when you 
want to measure two non-commuting variables simultaneously. 

If you would like to learn more about the uncertainty relation for the 
joint measurement of non-commuting variables you may have a look at 
the pedagogical article by M.G. Raymer, Am. J. Phys. 62, 986 (1994) 
which you can find in the library. 

2.3.1 The trace of an operator 

In the fourth postulate I have defined the rule for the calculation of 
the probability that a measurement result is one of the eigenvalues of 
the corresponding operator. Now I would like to generalize this idea 
to situations in which we have some form of lack of knowledge about 
the measurement outcome. An example would be that we are not quite 
sure which measurement outcome we have (the display of the apparatus 
may have a defect for example). To give this law a simple formulation, 
I need to introduce a new mathematical operation. 

Definition 39 The trace of an operator A on an N dimensional Hilbert 
space is defined as 


tr(A) = J2(Hm) (2-20) 

for any orthonormal set of basis vectors 

In particular, if the are chosen as the eigenvectors of A we see 
that the trace is just the sum of all the eigenvalues of A. The trace will 
play a very important role in quantum mechanics which we will see in 
our first discussion of the measurement process in quantum mechanics. 



The trace has quite a few properties that are helpful in calculations. 
Note that the trace is only well defined, if it is independent of the choice 
of the orthonormal basis {iV'i)}- If we chose a different basis {| </>»)}, then 
we know that there is a unitary operator U such that \ipi) = U\4>i) and 
we want the property tr{A} = tr{wAU}. That this is indeed true 

Theorem 40 For any two operators A and Bona Hilbert space H 
and unitary operators U we have 

tr{AB} = tr{BA} (2.21) 
tr{A} = tr{UAU ] } . (2.22) 

Proof: I prove only the first statement, because the second one follows 
directly from WU = 1 and Eq. (2.21). The proof runs as follows 


tr{AB} = £<^ n |i% n ) 



n=l k=l 
N N 

k=l n=l 

= J2J2WB\ij n )(ij n \A\ij k ) 

k=l n=l 
N N 
= J2(^\Bj2\^n)(^n\A\ij k ) 
k=l n=l 




= J2MBA\ip k ) 


= tr{BA} . 



Comment: If you write an operator in matrix form then the trace is 
just the sum of the diagonal elements of the matrix. 

I introduced the trace in order to write some of the expressions from 
the previous chapters in a different form which then allow their gener- 
alization to situations which involve classical uncertainty. 

In postulate 4 the probability to find eigenvalue a* in a measurement 
of the observable A has been given for a system in quantum state \ip). 
Using the trace it can now be written 

Pi = ^|Pi|^>=*r{A|^>^|} • 

Using the trace, the expectation value of an observable A measured 
on a system in a quantum state \ip) can be written as 

(A) w = ^\A\^} = tr{A\^m . 
2.4 The density operator 

When discussing measurement theory it is quite natural to introduce 
a new concept for the description of the state of quantum system. So 
far we have always described a quantum system by a state vector, i.e. 
by a pure state which may be a coherent superposition of many other 
state vectors, e.g. 

|^) = ai|^i) + a 2 |^2) . (2.23) 

A quantum mechanical system in such a state is ideal for exhibiting 
quantum mechanical interference. However, it has to be said that such 
a state is an idealization as compared to the true state of a quantum 
system that you can prepare in an experiment. In an experiment you 
will always have some statistical uncertainty due to imperfections in 
the preparation procedure of the quantum state or the measurement. 
For example, due to an occasional error, happening at a random time 
and unknown to you, the wrong quantum state may be prepared. Then 
we do not only have the quantum mechanical uncertainty that is de- 
scribed by the state vector, but we also have a classical statistical 
uncertainty. This raises the question as to how to describe such an 



Figure 2.1: An oven emits atomic two-level systems. The internal state 
of the system is randomly distributed. With probability pi the system 
is in the pure state \ipi). A person oblivious to this random distribution 
measures observable A. What is the mean value that he obtains? 

experimental situation in the most elegant way. Certainly it cannot 
be dealt with by just adding the state vectors together, i.e. forming a 
coherent superposition. 

To understand this better and to clarify the definition of the density 
operator I will present an example of an experimental situation where 
the description using pure quantum states fails, or better, where it is 
rather clumsy. Consider the situation presented in Fig. 2.1. An oven 
is filled with atomic two-level systems. We assume that each of these 
two-level systems is in a random pure state. To be more precise, let 
us assume that with probability Pi a two-level system is in state 
Imagine now that the oven has a little hole in it through which the 
two-level atoms can escape the oven in the form of an atomic beam. 
This atomic beam is travelling into a measurement apparatus built by 
an experimentalist who would like to measure the observable A. The 
task of a theoretical physicist is to predict what the experimentalist 
will measure in his experiment. 

We realize that each atom in the atomic beam is in a pure state 
with probability pi. For each individual atom it is unknown to the 
experimentalist in which particular state it is. He only knows the prob- 
ability distribution of the possible states. If the experimentalist makes 
measurements on N of the atoms in the beam, then he will perform 
the measurement JVj Npi times on an atom in state l^j). For each of 
these pure states \ipi) we know how to calculate the expectation value 
of the observable A that the experimentalist is measuring. It is simply 
(tpi\A\ipi) = tr{A\ifji){ifji\}. What average value will the experimentalist 
see in N measurements? For a large N the relative frequencies of oc- 



currence of state is Ni/N = p^. Therefore the mean value observed 
by the experimentalist is 

(i) = £^r{i|^}<^|} • (2-24) 


This equation is perfectly correct and we are able to calculate the ex- 
pectation value of any observable A for any set of states (iV'i)} an d 
probabilities {pi}- However, when the number of possible states \ipi) is 
really large then we have a lot of calculations to do. For each state 
we have to calculate the expectation value (ipi\A\ipi) = ir{A|^j)(^j|} 
and then sum up all the expectation values with their probabilities. If 
we then want to measure a different observable B then we need to do 
the same lengthy calculation again. That's not efficient at all! As a 
theoretical physicist I am of course searching for a better way to do 
this calculation. Therefore let us reconsider Eq. (2.24), transforming 
it a bit and use it to define the density operator. 

(A) = Y,pMMA)(A\} ■ 


= tr{AY,Pi\A)(A\} 

=: tr{Ap} . (2.25) 

The last equality is really the definition of the density operator p. 
Quite obviously, if I know the density operator p for the specific situa- 
tion - here the oven generating the atomic beam - then I can calculate 
quite easily the expectation value that the experimentalist will mea- 
sure. If the experimentalist changes his apparatus and now measures 
the observable B then this is not such a big deal anymore. We just need 
to re-evaluate Eq. (2.25) replacing A by B and not the potentially huge 
number of expectation values tr^lipj)^]}. 

Exercise: Check that in general we have (f(A)) = tr{f(A)p}. 

Before I present some useful properties of a density operator, let 
me stress again the difference between a coherent superposition of 
quantum states and a statistical mixture of quantum states. In a 
coherent superposition state of a quantum mechanical system, each 



representative of the ensemble is in the same pure state of the form 

|V) = £c^> , (2.26) 


which can be written as the density operator 

P=I^)(^I=E«^I^)(^I- (2-27) 

This is completely different to a statistical mixture where a repre- 
sentative is with a certain probability \ai\ 2 in a pure state The 
corresponding density operator is 

p = ]TH 2 |^>(^| • (2.28) 


and is also called a mixed state. You should convince yourself that 
indeed the two states Eqs. (2.26-2.28) are different, i.e. 

p = E|a J | 2 |^)(^l^l^)(^l • (2-29) 

Now let us consider some properties of the density operator which 
can in fact be used as an alternative definition of the density operator. 

Theorem 41 Any density operator satisfies 

1. p is a Hermitean operator. 

2. p is a positive semidefinite operator, i.e. V|^) : (ip\p\ip) > 0. 

3. tr{p} = 1 

Proof: As we have defined the density operator already we need to 
proof the above theorem. From our definition of the density operator 
we find 

P j = (EP<l^)^l) t = E^iX^I = P ■ (2.30) 

i i 

This followed because \ipi){ipi\ is a projection operator. This proves 
part 1). Part 2) follows by 

(V>|/#> = WKEpMWM = £p*<V#*><^> = I>|<v#*>l 2 > o . 

i i i 




The last property follows from the fact that probabilities are normal- 
ized, i.e. J2iPi — 1- Then we can see that 

Hp} = tr{Y,Pi\A)(A\) = Y.pM\i>i)(A\} = = 1 • ( 2 - 32 ) 

This finishes the proof □. 

Some simple but useful properties that can be derived from Theorem 
41 are 

2. All eigenvalues of p lie in the interval [0, 1]. 
Proof: Exercise! 

A natural question that arises when one considers the density oper- 
ator, is that of its decomposition into pure states, i.e. p = J^iPil^i) (^il- 
ls this decomposition unique, or are there many possible ways to obtain 
a given density operator as a statistical mixture of pure states? The 
answer can be found by looking at a particularly simple example. Let 
us consider the density operator representing the 'completely mixed' 
state of a two-level system, i.e. 

Looking at Eq. (2.33) we readily conclude that this density operator 
can be generated by an oven sending out atoms in states |0) or |1) with 
a probability of 50% each, see part a) of Fig . (2.2). That conclusion is 
perfectly correct, however, we could also imagine an oven that generates 
atoms in states |±) = (|0) ± |l))/\/2 with a probability of 50% each. 
Let us check that this indeed gives rise to the same density operator. 

p = ^(l+X+l + l-X-l) 

= |(|0><0| + |1)(0| + |0)(1| + |1)(1|) + |(|0><0| - |1)(0| - |0)(1| + |1)(1|) 

P = 

^(|0)(0| + |1)(1|) . 


2(|0)(0| + |1)(1|) • 




Figure 2.2: a) An oven generates particles in states |0) or |1) with 
probability 50%. An experimentalist measures observable A and finds 
the mean value tr{Ap}. b) The oven generates particles in states |±) = 
(|0) ± |l))/y / 2 with probability 50%. The experimentalist will find the 
same mean value tr{Ap}. 

Therefore the same density operator can be obtained in different 
ways. In fact, we can find usually find infinitely many ways of gen- 
erating the same density operator. What does this mean? Does this 
make the density operator an ill-defined object, as it cannot distinguish 
between different experimental realisations? The answer to that is a 
clear NO! The point is, that if a two realisations give rise to the same 
density operator, then, as we have seen above, we will also find exactly 
the same expectation values for any observable that we may chose to 
measure. If a quantum system in a mixed state gives rise to exactly the 
same predictions for all possible measurements, then we are forced to 
say that the two states are indeed the same! Of course this is something 
that needs to be confirmed experimentally and it turns out that it is 
true. Even more interestingly it turns out that if I could distinguish 
two situations that are described by the same density operator then I 
would be able to transmit signals faster than light! A clear impossi- 
bility as it contradicts special relativity. This is what I will explain to 
you in the next section using the concept of entanglement which you 
are going to encounter for the first time. 


2.5 Mixed states, Entanglement and the 
speed of light 

In the previous section I have introduced the density operator to de- 
scribe situations in which we have a lack of knowledge due to an 
imperfect preparation of a quantum state. However, this is not the 
only situation in which we have to use a density operator. In this 
section I will show you that the density operator description may also 
become necessary when the system that we are investigating is only the 
accessible part of some larger total system. Even if this larger 
system is in a pure quantum state, we will see that the smaller system 
behaves in every respect like a mixed state and has to be described by 
a density operator. This idea will then lead us to the insight that two 
different preparations that are described by the same density operator 
can not be distinguished experimentally, because otherwise we would 
be able to send messages faster than the speed of light. This would 
clearly violate the special theory of relativity. 

To understand this new way of looking at density operators, I first 
need to explain how to describe quantum mechanical systems that con- 
sist of more than one particle. This is the purpose of the next subsec- 

2.5.1 Quantum mechanics for many particles 

In this section I will show you what you have to do when you want to 
describe a system of two particles quantum mechanically. The general- 
ization to arbitrary numbers of particles will then be obvious. 

If we have two particles, then the state of each particle are elements 
of a Hilbert space; for particle A we have the Hilbert space T-La which 
is spanned by a set of basis states {\(f>i)A}i=i,...,N, while particle B has 
the Hilbert space TLb spanned by the set of basis states {\ipj)B}j=i,...,M- 
The two Hilbert spaces are not necessarily equal and may describe total 
different quantities or particles. 

Now imagine that system A is in state \(f>i)A and system B is in 
state \i])j)B- Then we write the total state of both systems in the tensor 


product form 

\*u*) = 10.) ® ■ (2.35) 

The symbol ® denotes the tensor product, which is not to be confused 
with ordinary products. Clearly there are N • M such combinations 
of basis vectors. These vectors can be thought of as spanning a larger 
Hilbert space Hab = <8> "Hb which describes all possible states that 
the two particles can be in. 

Of course we have to be able to add different states in the new 
Hilbert space Hab- We have to define this addition such that it is 
compatible with physics. Imagine that system A is in a superposition 
state \&)a = + 02(02) and system B is in the basis state |^) = 

|^i ). Then 

|$>a ® |#> = (ai|0i>A + a 2 |0 2 ) A ) ® (2.36) 

is the joint state of the two particles. Now surely it makes sense to say 

|$) A <8> = a^A® + a 2 |0 2 ) A ® IV'i) • (2.37) 

This is so, because the existence of an additional particle (maybe at the 
end of the universe) cannot change the linearity of quantum mechanics 
of system A. As the same argument applies for system B we should 
have in general 

(l>|<^ ® ^5>l^>) = E%"l&> ® l^> • (2-38) 

The set of states { 1 0j ) a ® I^^b} forms a basis of the state space of two 

Now let us see how the scalar product between two states <8> 
IV^b and <8> |0 2 )s has to be defined. Clearly if \tp 2 ) = 102) then 
we should have 

(\iPi)a ® \iI> 2 )b, \<Pi)a ® |0 2 )b) = (I^i)a, |0i)a) (2.39) 

again because the existence of an unrelated particle somewhere in the 
world should not change the scalar product of the first state. 



Eq. (2.39) can only be true if in general 

(|^i)a®|^2)bJ^i)a®|02)b) = (|^i)a,|^i)a)(|V'2)b,|02)b) • (2.40) 

The tensor product of linear operators is defined by the action of the 
operator on all possible product states. We define 

A <g) B(\<j>) A ® |V)b) := (A\<j>) A ) ® (%) B ) . (2.41) 

In all these definitions 1 have only used states of the form |0) <g) \ip) 
which are called product states. Are all states of this form? The 
answer is evidently no, because we can form linear superpositions of 
different product states. An example of such a linear superposition is 
the state 

|*) = -L(|00> + |11)) . (2.42) 

Note that 1 now omit the ® and abbreviate |0) (g> |0) by |00). If you try 
to write the state Eq. (2.42) as a product 

HI = (a|0)+/3|l»<g>(7|0)+<?|l» = a 7 |00)+a5|01)+/3 7 |10)+^|ll) 


This implies that aS = = (3^. If we chose a = then also cry = and 
the two sides cannot be equal. Likewise if we chose 5 = then (35 = 
and the two sides are different. As either a = or 5 = we arrive at 
a contradiction. States which cannot be written in product form, are 
called entangled states. In the last part of these lectures we will learn 
a bit more about the weird properties of the entangled states. They al- 
low such things as quantum state teleportation, quantum cryptography 
and quantum computation. 

Let us now see how the tensor product looks like when you write 
them as column vectors? The way you write them is really arbitrary, 
but of course there are clumsy ways and there are smart ways. In 
literature you will find only one way. This particular notation for the 
tensor product of column vectors and matrices ensures that the relation 
A<S>B\</>) <S> = A\<f>) <S>B\ip) is true when you do all the manipulations 


using matrices and column vectors. The tensor product between a Tri- 
dimensional vector and an n-dimensional vector is written as 

( ^ \ 

\a m J 

( ^ \ 

\b n J 

f ( b i \ \ 

o-i : 
V b n ) 


( bl \ 

\b n J 

1 ai&i \ 
V a m bn J 



The tensor product of a linear operator A on an m-dimensional space 
and linear operator B on an n-dimensional space in matrix notation is 
given by 

' an 

\ 0>ml 


dim \ ( bu 


' fen ... bi n \ 

\ b n \ ■ ■ ■ b nn j 

bin \ 
bnn J 


( a n B 

( b 




hn \ 

V fenl 

\ a-miB . . . a mm B J 

' fell • • • feln \ \ 
\ fenl • • • b nn J 
/fell • • • feln \ 




. (2.44) 

The simplest explicit examples that I can give are those for two 2- 
dimensional Hilbert spaces. There we have 

a 2 

fe 2 

( aifei ^ 
aife 2 
a 2 fei 
V a 2 fe 2 J 





Oil Ol2 
a 21 °22 

hi bi2 

^21 &22 

/ Oll&ll 

a 2 i&n 

V 021&21 



ai 2 6n 

a 22^21 

a 22^22 


2.5.2 How to describe a subsystem of some large 

Let us now imagine the following situation. The total system that 
we consider consists of two particles each of which are described by an 
N(M)-dimensional Hilbert space Ha (Hb) with basis vectors {\4>i) a}i=i,...,n 

b}i=i,...,m) ■ Let us also assume that the two particles together are 
in an arbitrary, possibly mixed, state pab- One person, Alice, is holding 
particle A, while particle B is held by a different person (let's call him 
Bob) which, for whatever reason, refuses Alice access to his particle. 
This situation is schematically represented in Fig. 2.3. Now imagine 
that Alice makes a measurement on her system alone (she cannot ac- 
cess Bobs system). This means that she is measuring an operator of 
the form A <g) 1. The operator on Bobs system must be the identity 
operator because no measurement is performed there. I would now like 
to calculate the expectation value of this measurement, i.e. 

(A) = ^ A ((f>\B(ip j \A®l\(f>) A \ip j ) = tr A B{A®lp A B} , 


where trAB means the trace over both systems, that of Alice and 
that of Bob. This description is perfectly ok and always yields the 
correct result. However, if we only ever want to know about outcomes 
of measurements on Alice's system alone then it should be sufficient to 
have a quantity that describes the state of her system alone without 
any reference to Bob's system. I will now derive such a quantity, which 
is called the reduced density operator. 

Our task is the definition of a density operator pa for Alices system 
which satisfies 

{A) = trAB {A ® Ipab} = tr A {Ap A } (2.47) 


Figure 2.3: Alice and Bob hold a joint system, here composed of two 
particles. However, Alice does not have access to Bob's particle and 
vice versa. 

for all observables A. How can we construct this operator? Let us 
rewrite the left hand side of Eq. (2.47) 

N M 
i=l 3=1 

N I M \ 

= £a<0*|^. \^B{^j\pAB\^j)B )\<f>i)A 
i=l \j=l J 

- ( M \ 
= tr A {A I Y^B^PAB^jiB 1} • 

In the last step I have effectively split the trace operation into two parts. 
This split makes it clear that the state of Alices system is described by 



the reduced density operator 


pA-=Y.B^j\pAB\^j)B=tr B {p AB } . (2.48) 


where the last identity describes the operation of taking the partial 
trace. The point is now that the reduced density operator allows Alice 
to describe the outcome of measurements on her particle without any 
reference to Bob's system, e.g. 

(A) = tr A {A PA } . 

When I am giving you a density operator p AB then you will ask 
how to actually calculate the reduced density operator explicitly. Let 
me first consider the case where p AB is the pure product state, i.e. 
Pab = \4>)a\4>i)ba{<P\b(^i\ Then following Eq. (2.48) the partial trace 


tr B {\(f)M((f)M} = ^B{A\{\Mi)((t>M)\A)B 



= J2\^)^A(<Pl\B(^i\^l)BB(^l\^i)B 

= |0i)aa(0i| • (2.49) 

Now it is clear how to proceed for an arbitrary density operator of the 
two systems. Just write down the density operator in a product basis 

N M 

PAB = E UiiM^ijAA^M ® \4>j)BB(H , (2.50) 

i,fc=l j,l=l 

and then use Eq. (2.49) to find 

m I n m \ 

tr B {p AB } = ^2b(^c\ E 2 AA((pk\ <S> \i>j)BB(lpl\ \iPc)b 

c=l \i,k=lj,l=l J 

M N M 

= EE E a ijM \(t)i) AA ((t) k \{B{^Mj)B){B{i'l\^c)B) 
c=l i,k=l j,l=l 


M N M 

= a ij,kl\4>i) AA((f>k\S c jSlc 

c=l i,k=l j,l=l 


Let me illustrate this with a concrete example. 
Example: Consider two system A and B that have two energy lev- 
els each. I denote the basis states in both Hilbert spaces by |1) and 
1 2). The basis states in the tensor product if both spaces are then 
(|11), 1 12), 1 21), 1 22)}. Using this basis I write a possible state of the 
two systems as 

Pab = ||11><H| + ||11><12| + ||11><21| 

+ i|12)(ll| + i|12)(12| + i|12)(21| 

+ g|21)(ll| + g|21)(12| + g|21)(21| . (2.52) 

Now let us take the partial trace over the second system B. This means 
that we collect those terms which contain states of the form \i)bb{A- 
Then we find that 

PA = §|1><1| + ^|1)(2| + +^|2)(1| + ±|2><2| . (2.53) 

This is not a pure state (Check that indeed detp A ^ 0). It is not 
an accident that I have chosen this example, because it shows that the 
state of Alice's system may be a mixed state (incoherent superposition) 
although the total system that Alice and Bob are holding is in a pure 
state. To see this, just check that 

Pab = \V)abab(*\ , (2.54) 

where |*)ab = \Z||H) + \Z|I 12 ) + \/§|21). This is evidently a pure 
state! Nevertheless, the reduced density operator Eq. (2.53) of Alice's 
system alone is described by a mixed state! 

In fact this is a very general behaviour. Whenever Alice and Bob 
hold a system that is in a joint pure state but that cannot be written 



as a product state \<f>) <g> \ip) (Check this for Eq. (2.54), then the reduced 
density operator describing one of the subsystems represents a mixed 
state. Such states are called entangled states and they have quite a lot 
of weird properties some of which you will encounter in this lecture and 
the exercises. 

Now let us see how these ideas help us to reveal a connection between 
special relativity and mixed states. 

2.5.3 The speed of light and mixed states. 

In the previous subsection you have seen that a mixed state of your 
system can also arise because your system is part of some larger inac- 
cessible system. Usually the subsystem that you are holding is then in 
a mixed state described by a density operator. When I introduced the 
density operator, I told you that a particular mixed state be realized 
in different ways. An example was 

with |±) = ^ (| 1) ± 1 2)). However, I pointed out that these two real- 
izations are physically the same because you are unable to distinguish 
them. I did not prove this statement to you at that point. I will now 
correct this omission. I will show that if you were able to distinguish 
two realizations of the same density operator, then you could send sig- 
nals faster than the speed of light, i.e. this would violate special 
relativity. How would that work? 

Assume two persons, Alice and Bob, hold pairs of particles where 
each particle is described by two-dimensional Hilbert spaces. Let us 
assume that Alice and Bob are sitting close together, and are able to 
access each others particle such that they are able to prepare the state 

Now Alice and Bob walk away from each other until they are separated 
by, say, one light year. Now Alice discovers the answer to a very tricky 


2|l)(l| + 2l 2 )( 2 l 



question which is either 'yes' or 'no' and would like to send this answer 
to Bob. To do this she performs either of the following measurements 
on her particle. 

'yes' Alice measures the observable A yes = |1)(1| + 2|2)(2|. 

The probability to find the state |1) is p\ = \ and to find |2) is 
p 2 = \. This follows from postulate 4 because Pi = tr{\i){i\p A } = 
| where pa = |(|1)(1| + |2)(2|) is the reduced density operator 
describing Alices system. After the result |1) the state of the total 
system is |1)(1| <8> lL]-0) ~ |11), after the result |2) the state of the 
total system is |2)(2| ® t\ip) ~ |22). 

'no' Alice measures the observable A no = |+)(+| + 2|— )( — | ^ A yes . 
The probability to find the state |+) is p + — | and to find |— ) is 
P- = \- This follows from postulate 4 because Pi = tr{\i)(i\pA} = 
I where pa = |(|+)(+| + |— )(— I) is the reduced density operator 
describing Alices system. After the result |+) the state of the 
total system is |+)(+| <8> l\ip) ~ | + +), after the result |— ) the 
state of the total system is |— )(— | <8> l\ip) ~ | ). 

What does this imply for Bob's system? As the measurement results 
1 and 2 occur with probability 50%, Bob holds an equal mixture of 
the two states corresponding the these outcomes. In fact, when Al- 
ice wanted to send 'yes' then the state of Bob's particle after Alices 
measurement is given by 

P^ = ^|0)(0| + ^|1)(1| (2.56) 

and if Alice wanted to send 'no' then the state of Bob's particle is given 
by : | 

Pno=\\+){+\ + \\-){-\ ■ (2-57) 

You can check quite easily that p no = p yes . If Bob would be able to 
distinguish the two realizations of the density operators then he could 
find out which of the two measurements Alice has carried out. In that 
case he would then be able to infer what Alices answer is. He can do 
that immediately after Alice has carried out her measurements, which 
in principle requires negligible time. This would imply that Alice could 



send information to Bob over a distance of a light year in virtually no 
time at all. This clearly violates the theory of special relativity. 

As we know that special relativity is an extremely well established 
and confirmed theory, this shows that Bob is unable to distinguish the 
two density operators Eqs. (2.56-2.57). 

2.6 Generalized measurements 

Chapter 3 

Dynamics and Symmetries 

So far we have only been dealing with stationary systems, i.e. we did 
not have any prescription for the time evolution of quantum mechan- 
ical systems. We were only preparing systems in particular quantum 
mechanical states and then subjected them to measurements. In real- 
ity of course any system evolves in time and we need to see what the 
quantum mechanical rules for time evolution are. This is the subject 
of this chapter. 

Most of you will know that the prescription that determines the quan- 
tum mechanical time evolution is given by the Schrodinger equation. 
Before I state this as the next postulate, let us consider what any de- 
cent quantum mechanical time evolution operator should satisfy, and 
then we will convince us that the Schrodinger equation satisfies these 
criteria. In fact, once we have accepted these properties, we will not 
have too much freedom of choice for the form of the Schrodinger equa- 

Definition 42 The time evolution operator U (t 2 , t\) maps the quantum 
state \i])(t\)) to the state \1pit2)) obeying the properties 

1. U(t2,ti) is unitary. 

3.1 The Schrodinger Equation 





We have U(t2,ti)U(ti,to) = Ufa, to), (semi-group property) 


U(t,ti) is differentiable in t. 

What is the reason for these assumptions? First of all, the time 
evolution operator has to map physical states onto physical states. That 
means in particular that a normalized state \i/j(to)) is mapped into a 
normalized state \ip(ti)), i.e. for any \ip(t )) 

The second property of the time evolution operator demands that 
it does not make a difference if we first evolve the system from t to 
t\ and then from t\ to £2 or if we evolve it directly from time to to £2- 
This is a very reasonable assumption. Note however, that this does not 
imply that we may measure the system at the intermediate time t\. In 
fact we must not interact with the system. 

The third condition is one of mathematical convenience and of phys- 
ical experience. Every system evolves continuously in time. This is 
an observation that is confirmed in experiments. Indeed dynamics in 
physics is usually described by differential equations, which already im- 
plies that observable quantities and physical state are differentiable in 
time. You may then wonder what all the fuss about quantum jumps 
is then about. The point is that we are talking about the time evo- 
lution of a closed quantum mechanical system. This means that we, 
e.g. a person from outside the system, do not interact with the system 
and in particular this implies that we are not measuring the system. 
We have seen that in a measurement indeed the state of a system can 
change discontinuously, at least according to our experimentally well 
tested Postulate 4. In summary, the quantum state of a closed system 
changes smoothly unless the system is subjected to a measurement, i.e. 
an interaction with an outside observer. 

What further conclusions can we draw from the properties of the 
time evolution operator? Let us consider the time evolution opera- 
tor for very short time differences between t and t and use that it is 



differentiable in time (property 3) of its Definition 42). Then we find 

U(t,t ) = l-^H(t )(t-t ) + ... . (3.2) 

Here we have used the fact that we have assumed that the time evolu- 
tion of a closed quantum mechanical system is smooth. The operator 
if (to) that appears on the right hand side of Eq. (3.2) is called the 
Hamilton operator of the system. Let us apply this to an initial 
state \ip(t )) and take the time derivative with respect to t on both 
sides. Then we find 

ihdtMt)) = ihd t (U(t,t )m ))) » H(toM(t )) . (3.3) 

If we now carry out the limit t — > to then we finally find 

m\m) = H(t)\m) ■ (3-4) 

This is the Schrodinger equation in the Dirac notation in a coordinate 
independent form. To make the connection to the Schrodinger equation 
as you know it from last years quantum mechanics course let us consider 
Example: Use the Hamilton operator for a particle moving in a one- 
dimensional potential V(x,t). Write the Schrodinger equation in the 
position representation. This gives 

ihdt(x\iP(t)) = (x\ihd t \i;(t)) 
= (x\HW)) 

This is the Schrodinger equation in the position representation as you 
know it from the second year course. 

Therefore, from the assumptions on the properties of the time evolu- 
tion operator that we made above we were led to a differential equation 
for the time evolution of a state vector. Again, of course, this result 
has to be tested in experiments, and a (sometimes) difficult task is to 
find the correct Hamilton operator H that governs the time evolution 
of the system. Now let us formulate the result of our considerations in 



the following 

Postulate 5 The time evolution of the quantum state of an isolated 
quantum mechanical system is determined by the Schrddinger equation 

m\m) = H(t)\m) > (3.5) 

where H is the Hamilton operator of the system. 

As you may remember from your second year course, the Hamilton 
operator is the observable that determines the energy eigenvalues of the 
quantum mechanical system. In the present course I cannot justify this 
in more detail, as we want to learn a lot more things about quantum 
mechanics. The following discussion will however shed some more light 
on the time evolution of a quantum mechanical system and in particular 
it will pave our way towards the investigation of quantum mechanical 
symmetries and conserved quantities. 

3.1.1 The Heisenberg picture 

In the preceding section we have considered the time evolution of the 
quantum mechanical state vector. In fact, we have put all the time evo- 
lution into the state and have left the observables time independent. 
This way of looking at the time evolution is called the Schrddinger 
picture. This is not the only way to describe quantum dynamics (un- 
less they had some intrinsic time dependence). We can also go to the 
other extreme, namely leave the states time invariant and evolve the 
observables in time. How can we find out what the time evolution of 
a quantum mechanical observable is? We need some guidance, some 
property that must be the same in both pictures of quantum mechan- 
ics. Such a property must be experimentally observable. What I will 
be using here is the fact that expectation values of any observable has 
to be the same in both pictures. Let us start by considering the expec- 
tation value of an operator As in the Schrddinger picture at time time 



t given an initial state \ip(to)). We find 

(A) = (^(t)\A s \m) = m )\u%t )A s u(t,t )m )) . (3.6) 

Looking at Eq. (3.6) we see that we can interpret the right hand side 
also as the expectation value of a time-dependent operator 

A H (t) = tf(t,t )A s U(t,t ) (3.7) 
in the initial state \ip(t )). That is 

(m\Mm) = = m )\A H (t)\4>(t )) . m 

Viewing the state of the system as time-independent, and the observ- 
ables as evolving in time is called the Heisenberg-picture of quantum 
mechanics. As we have seen in Eq. (3.8) both, the Schrodinger picture 
and the Heisenberg picture, give exactly the same predictions for physi- 
cal observables. But, depending on the problem, it can be advantageous 
to chose one or the other picture. 

Like the states in the Schrodinger picture the Heisenberg operators 
obey a differential equation. This can easily be obtained by taking the 
time derivative of Eq. (3.7). To see this we first need to know the time 
derivative of the time evolution operator. Using \ip(t)) = U (t, to) (to) ) 
in the Schrodinger equation Eq. (3.4) we find for any \ip(to)) 

ihd t U(t,t )\i>(t )) = H(t)U(t,t )\i>(t )) (3.9) 

and therefore 

ihd t U(t,t ) = H(t)U(t,t ) . (3.10) 

Assuming that the the operator in the Schrodinger picture has no 
explicit time dependence, we find 

j f ((/*((, t a )A s U(t,t )) 
^|«i s( 7( Mo ) + fr t (Mo)is ^|« 

l -U\t,t„)H{t)A s U{t,t ) + UHt,t a )A s l -H(t)U(t,t„) 



l -U\t,t )[H(t),A s ]U(t,t ) 

lm),A s ] H 

l -[H H (t),A H ] . 



It is easy to check that for an operator that has an explicit time depen- 
dence in the Schrodinger picture, we find the Heisenberg equation 

One of the advantages of the Heisenberg equation is that it has a direct 
analogue in classical mechanics. In fact this analogy can be viewed as 
a justification to identify the Hamiltonian of the system H with the 
energy of the system. I will not go into details here and for those of 
you who would like to know more about that I rather recommend the 
book: H. Goldstein, Classical Mechanics, Addison- Wesley (1980). 

3.2 Symmetries and Conservation Laws 

After this introduction to the Heisenberg picture it is now time to 
discuss the concept of symmetries and their relation to conservation 

3.2.1 The concept of symmetry 

If we are able to look at a system in different ways and it appears the 
same to us then we say that a system has a symmetry. This simple 
statement will be put on solid ground in this section. Why are we 
interested in symmetries? The answer is simple! It makes our life 
easier. If we have a symmetry in our problem, then the solution will 
also have such a symmetry. This symmetry of the solution expresses 
itself in the existence of a quantity which is conserved for any solution 
to the problem. Of course we are always interested to know quantities 
that are conserved and the natural question is how these conserved 
quantities are related to the symmetry of the problem! There are two 




ways to formulate symmetries, in an active way and a passive way. 
If we assume that we transform the system S into a new system S' 
then we speak of an active transformation, i.e. we actively change 
the system. If we change the coordinate system in which we consider 
the same system, then we speak of a passive transformation. In these 
lectures I will adopt the active point of view because it appears to be the 
more natural way of speaking. An extreme example is the time reversal 
symmetry. Certainly we can make a system in which all momenta of 
particles are reversed (the active transformation) while we will not be 
able to actually reverse time, which would amount to the corresponding 
passive transformation. 

Now let us formalize what we mean by a symmetry. In the active 
viewpoint, we mean that the systems S and S' = TS look the same. We 
have to make sure what we mean by J S and S' look the same'. What 
does this mean in quantum mechanics? The only quantities that are 
accessible in quantum mechanics are expectation values or, even more 
basic, transition probabilities. Two systems are the same in quantum 
mechanics if the transition probabilities between corresponding states 
are the same. To be more precise let us adopt the 

Definition 43 A transformation T is called a symmetry transfor- 
mation when for all vectors \<p) and in system S and the corre- 
sponding vectors \(f>') = T\<f>) and \ip') = T\ip) in system S' we find 

IW)I = IW)I • (3-14) 

In this definition we have seen that quantum states are transformed 
as = T\(f). This implies that observables, which are of the form 
A = Y^i ai\ai) (ai\ will be transformed according to the prescription 
A' = fAfl 

Which transformations T are possible symmetry transformations? 
As all transition probabilities have to be preserved one would expect 
that symmetry transformations are automatically unitary transforma- 
tions. But this conclusion would be premature because of two reasons. 
Firstly, we did not demand that the scalar product is preserved, but 
only the absolute value of the scalar product and secondly we did not 
even demand the linearity of the transformation T. Fortunately it turns 



out that a symmetry transformation T cannot be very much more gen- 
eral than a unitary transformation. This was proven in a remarkable 
theorem by Wigner which states 

Theorem 44 Any symmetry transformation can either be represented 
by a unitary transformation or an anti-unitary transformation. 

For a proof (which is rather lengthy) of this theorem you should 
have a look at books such as: K. Gottfried, Quantum Mechanics I, 
Benjamin (1966). 

An anti-unitary transformation has the property 

U(a\</>) + = a*U\(f)) + (3*U\ij) (3.15) 

and preserves all transition probabilities. An example of an anti-unitary 
symmetry is time-reversal, but this will be presented later. 

Of course a system will usually be symmetric under more than just 
one particular transformation. In fact it will be useful to consider 
symmetry groups. Such groups may be discrete (you can number them 
with whole numbers) or continuous (they will be parametrized by a real 
number) . 

Definition 45 A continuous symmetry group is a set of symmetry 
transformation that can be parametrized by a real parameter such that 
the symmetry transformations can be differentiated with respect to this 

Example: a) Any Hermitean operator A gives rise to a continuous 
symmetry group using the definition 

U(e) := e L ^ n . (3.16) 

Obviously the operator U (e) can differentiated with respect to the pa- 
rameter e. 

So far we have learned which transformations can be symmetry 
transformations. A symmetry transformation of a quantum mechanical 
system is either a unitary or anti-unitary transformation. However, 



for a given quantum system not every symmetry transformation is a 
symmetry of that quantum system. The reason for this is that it is not 
sufficient to demand that the configuration of a system is symmetric at 
a given instance in time, but also that this symmetry is preserved 
under the dynamics of the system! Otherwise we would actually be 
able to distinguish two 'symmetric' states by just waiting and letting 
the system evolve. The invariance of the configurational symmetry 
under a time evolution will lead to observables that are conserved in 
time. As we are now talking about the time evolution of a quantum 
system we realize that the Hamilton operator plays a major role in 
the theory of symmetries of quantum mechanical systems. In fact, this 
is not surprising because the initial state (which might posses some 
symmetry) and the Hamilton operator determine the future of a system 

Assume that we have a symmetry that is preserved in time. At the 
initial time the symmetry takes a state \ip) into = T\ip). The time 
evolution of the original system S is given by the Hamilton operator H 
while that of the transformed system S' is given by H' = THT^ . This 
leads to 

m) = e -iHt/H lm) (317) 

for the time evolution in S. The time evolution of the transformed 
system S' is governed by the Hamilton operator H' so that we find 

\^(t))' = e- iA ' t/n \i;(0)y . (3.18) 

However, the state \ip)' could also be viewed as quantum state of the 
original system 5! This means that the time evolution of system S 
given by the Hamilton operator H could be applied and we would find 

\^(t))" = e-' ifl ' tlTl \^))' . (3.19) 

If the symmetry of the system is preserved for all times then the two 
states \ip(t))' and \ip(t))" cannot differ by more than a phase factor, i.e. 

mt))" = J*\tKt)Y ■ (3-20) 

because we need to have 

\wm\ = \(m'\m")\ 




As Eq. (3.21) has to be true for all state vectors, the Hamilton operators 
H and H' can differ by only a constant which is physically unobservable 
(it gives rise to the same global e l(t> in all quantum states) and therefore 
the two Hamilton operators are essentially equal. Therefore we can 
conclude with 

Lemma 46 A symmetry of a system S that holds for all times needs to 
be a unitary or anti-unitary transformation T that leaves the Hamilton 
operator H invariant, i. e. 

H — H' — THT^ . (3.22) 

Given a symmetry group which can be parametrized by a parameter 
such that the symmetry transformations are differentiable with respect 
to this parameter (see the Example in this section above), we can de- 
termine the conserved observable G of the system quite easily from the 
following relation 

G = Um ^ (e) - 1 . (3.23) 

The quantity G is also called the generator of the symmetry group. 

Why is G conserved? From Eq. (3.23) we can see that for small e we 
can write 

17(e) = 1 - ieG + 0(e 2 ) , (3.24) 

where G is a Hermitean operator and 0(e 2 ) means that all other terms 
contain at least a factor of e 2 and are therefore very small, (see in the 
first chapter for the proof). Now we know that the Hamilton operator 
is invariant under the symmetry transformation, i.e. 

H = H' = U(e)HU\e) . (3.25) 

For small values of e we find 

H = (l-iGe)H(l + iGe) 

= H — i[G, H]e + 0(e 2 ) . (3.26) 

As the equality has to be true for arbitrary but small e, this implies 

[G,H] = . (3.27) 


Now we remember the Heisenberg equation Eq. (3.13) and realize that 
the rate of change of the observable G in the Heisenberg picture is 
proportional to the commutator Eq. (3.27), 

d ^i = i h[G,H] H = , (3.28) 

i.e. it vanishes. Therefore the expectation value of the observable G is 
constant, which amounts to say that G is a conserved quantity, i.e. 

(m\G\m) = (m\G H (t)\m) = _ {329) 

dt dt 
Therefore we have the important 

Theorem 47 The generator of a continuous symmetry group of a quan- 
tum system is a conserved quantity under the time evolution of that 
quantum system. 

After these abstract consideration let us now consider some exam- 
ples of symmetry groups. 

3.2.2 Translation Symmetry and momentum con- 

Let us now explore translations of quantum mechanical systems. Again 
we adopt the active viewpoint of the transformations. First we need to 
define what we mean by translating the system by a distance a to the 
right. Such a transformation T a has to map the state of the system \ip) 
into \ip a ) = T a \tp) such that 

(a# ) = (x- aW) . (3.30) 

In Fig. 3.1 you can see that this is indeed a shift of the wavefunction (i.e. 
the system) by a to the right. Which operator represents the translation 
operator T a ? To see this, we begin with the definition Eq. (3.30). Then 
we use the representation of the identity operator 1 = J dp\p){p\ (see 



Figure 3.1: The original wave function ip{x) (solid line) and the shifted 
wave function ip a (x) = ip(x — a). \ip a ) represents a system that has been 
shifted to the right by a. 

Eq. (1.131)) and Eq. (1.128). We find 

{x\f a \ijj) = (x-a\il>) 

j dp{x-a\p){p\ip) 

dp^=e ip{x ~ a),h {p\i)) 
dp—^e- ipa/n e ipx/n (p\ifj) 
= Jdpe~ ipa / n (x\p)(p\^) 

= (x\(J dpe- ipa ' h \p)(p\m 

= (x\e- tpa/h \4>) . (3.31) 

As this is true for any \tp), we that the translation operator has the 

f a = e - ipa/h . (3.32) 

This means that the momentum operator is the generator of transla- 
tions. From Eq. (3.23) it follows then that in a translation invariant 



system momentum is conserved, i.e. 

d(m\p\m) _ Q 


Consider the Hamilton operator of a particle in a potential, which 

H = £ + V(t) . (3.33) 

The momentum operator is obviously translation invariant, but a trans- 
lation invariant potential has to satisfy V(x) = V(x + a) for all a which 
means that it is a constant. 

3.2.3 Rotation Symmetry and angular momentum 

Another very important symmetry is rotation symmetry. To consider 
rotation symmetry it is necessary to go to three-dimensional systems. 
Let us first consider rotations R z (a) with an angle a around the z-axis. 
This is defined as 

Mr,(/>,9)=iP(r,<f>-a,e) . (3.34) 

Here we have written the wavefunction in polar coordinates 

x\ = r cos sin 6* (3.35) 
X2 = r sine/) sin # (3.36) 
x 5 = rcosd . (3.37) 

In Fig. 3.2 we can see that transformation Eq. (3.34) indeed rotates a 
wave-function by an angle a counter clockwise. 

As translations are generated by the momentum operator, we expect 
rotations to be generated by the angular momentum. Let us confirm 
our suspicion. The angular momentum of a particle is given by 

xxp (3.38) 
(x 2 p3 - x z p 2 )ei + (x 3 pi - xipz)e 2 + {x x p 2 - x 2 pi)e 3 (3.39) 



Figure 3.2: The original wave function i[>(r,(f),0) (solid line) and the 
rotated wave function ip a (r,(f),0) = i/)(r,<f> — a, 9). \ip a ) represents a 
system that has been rotated clockwise around the x 3 — axis. 

where x is the ordinary vector-product, p = f>\t\ + f> 2 e 2 + ^3^3 and 
x = x\i\ + x 2 e 2 + x 3 e 3 with [x i: pj] = ih5ij. 

As we are considering rotations around the z-axis, i.e. e*3, we will 
primarily be interested in the I3 component of the angular momentum 
operator. In the position representation we find 

This can be converted to polar coordinates by using the chain rule 

d dxi d dx 2 d 

d(p dxi d(j) dx 2 
d(r cos0sin#) d d(r sin0sin#) d 

d<fi dxi dcj) dx-2 

■ 1 ■ /> 9 , . , d 

—r sin smv- h r cos sin v- 

dxi dx 2 

d d 
-x 2 - h x 1 

dxi dx- z 



= \k . (3.41) 

Now we can proceed to find the operator that generates rotations by 
deriving a differential equation for it. 

, .. dRJa) , , . d , .. „ . . . , . 

= Q^{r,<l>-a,6\il)) 

= —Q^{r,<i>-a,6\il)) 

= ~{£\R,(a)\rl>) 

= ~(£\i 3 R,(a)\if>) . (3.42) 

As this is true for all wave functions \ip) we have the differential equation 
for the rotation operator 

= -n hR ^ a) (3 - 43) 

With the initial condition R z (0) = 1 the solution to this differential 
equation is given by 

R z (a) = e - liza/h = e - ltSia/h . (3.44) 
We obtain a rotation around an arbitrary axis n by 

R z ( a ) = e - iXaa ' n . (3.45) 

Any system that is invariant under arbitrary rotations around an 
axis n preserves the component of the angular momentum in that direc- 

— * 

tion, i.e. it preserves In. As the kinetic energy of a particle is invariant 
under any rotation (Check!) the angular momentum is preserved if the 
particle is in a potential that is symmetric under rotation around the 
axis n. The electron in a hydrogen atom has the Hamilton operator 



Rotations leave the length of a vector invariant and therefore \x\ and 
with it V(|x|) are invariant under rotations around any axis. This 
means that any component of the angular momentum operator is pre- 
served, i.e. the total angular momentum is preserved. 

3.3 General properties of angular momenta 

In the preceding section we have investigated some symmetry transfor- 
mations, the generators of these symmetry transformations and some 
of their properties. Symmetries are important and perhaps the most 
important of all is the concept of rotation symmetry and its generator, 
the angular momentum. In the previous section we have considered 
the specific example of the orbital angular momentum which we could 
develop from the classical angular momentum using the correspondence 
principal. However, there are manifestations of angular momentum in 
quantum mechanics that have no classical counterpart, the spin of an 
electron being the most important example. In the following I would 
like to develop the theory of the quantum mechanical angular momen- 
tum in general, introducing the notion of group representations. 

3.3.1 Rotations 

Whenever we aim to generalize a classical quantity to the quantum 
domain, we first need to investigate it carefully to find out those prop- 
erties that are most characteristic of it. Then we will use these basic 
properties as a definition which will guide us in our quest to find the cor- 
rect quantum mechanical operator. Correctness, of course, needs to be 
tested experimentally, because nice mathematics does not necessarily 
describe nature although the mathematics itself might be consistent. 

In the case of rotations I will not look at rotations about arbitrary 
angles, but rather at rotations about very small, infinitesimal angles. 
This will be most convenient in establishing some of the basic properties 
of rotations. 

As we have done in all our discussion of symmetries, we will adopt 
the viewpoint of active rotations, i.e. the system is rotated while the 
coordinate system remains unchanged. A rotation around an axis n 



for a positive angle is one that follows the right hand rule (thumb in 
direction of the axis of rotation). 

In three dimensional space rotations are given by real 3x3 matrices. 
A rotation around the x-axis with an angle is given by 


cos — sin 
sin cos 

n x {4>) 

The rotations around the y-axis is given by 


and the rotation about the z-axis is given by 

n z {<t>) 

( cos0 sin< 
y — sin0 cos< 

/ cos — sin \ 

sin cos 
V l) 

Using the expansions for sin and cos for small angles 




sine = e + 0(e 3 ) cose = l 


we may now expand the rotation matrices Eqs. (3.47-3.49) up to second 
order in the small angle e. We find 

TZ y (e) 

( 1 


— e 


V o 
/ i 


/ l- £ - 



e 2 




t 2 






We know already from experience, that rotations around the same axis 
commute, while rotations around different axis generally do not com- 
mute. This simple fact will allow us to derive the canonical commuta- 
tion relations between the angular momentum operators. 

To see the effect of the non-commutativity of rotations for different 
axis, we compute a special combination of rotations. First rotate the 
system about an infinitesimal angle e about the y-axis and then by the 
same angle about the x-axis 

K x (e)K y (e) 



— e 
1 - % 


V -e 






Now we calculate 

n x (-e)n y (-e)n x (e)n y (e) = 


( 1 " C 

e 2 1 

V o o 

n z {e 2 ) 

— e 
2 \ 

1 J 



— e 


This will be the relation that we will use to derive the commutation 
relation between the angular momentum operators. 

3.3.2 Group representations and angular momen- 
tum commutation relations 

We already know that there are many different angular momenta in 
quantum mechanics, orbital angular momentum and spin are just two 



examples. Certainly they cannot necessarily represented by 3 x 3 ma- 
trices as in the rotations in the previous section. Nevertheless, all these 
angular momenta have some common properties that justify their clas- 
sification as angular momenta. This will lead us to a discussion of 
groups and their representations. 

We have already seen in the first section on vector spaces what a 
group is. Let us repeat the properties of a group G and its elements 

1. Wx, y E G : x ■ y = z E G 

2. 31 G G : Wx G G : x ■ 1 = x 

3. Wx G G : 3x~ l G G : x^ 1 ■ x — 1 and a; • x" 1 = 1 

4. Vx, y, z G Gr : x • (y • z) — (x • y) • z 

Note that we did not demand that the group elements commute under 
the group operation. The reason is that we intend to investigate rota- 
tions which are surely not commutative. The fundamental group that 
we are concerned with is the group of rotations in three-dimensional 
space, i.e. the group of 3 x 3 matrices. It represents all our intuitive 
knowledge about rotations. Every other group of operators on wave- 
functions that we would like to call a group of rotations will have to 
share the basic properties of the group of three-dimensional rotations. 
This idea is captured in the notion of a group representation. 

Definition 48 A representation of a group G of elements x is a group 
S of unitary operators V(x) such that there is a map T : G — > S 
associating with every x G G an operator T>(x) G S such that the group 
operation is preserved, i.e. for all x,y G G with x ■ y = z we have 

This means that both groups G and S are essentially equal because 
their elements share the same relations between each other. If we now 
have a representation of the rotation group then it is natural to say that 
the operators that make up this representation are rotation operators. 




In quantum mechanics a rotation has to be represented by a unitary 
operator acting on a state vector. For rotations around a given axis n 
these unitary operators have to form a set of operators U((f>) that is 
parametrized by a single parameter, the angle 0. As we know that the 
operators are unitary we know that they have to be of the form 

= e -<-W/» 

with a Hermitean operator J n and where we have introduced % for 
convenience. The operator is the angular momentum operator along 
direction n. In general we can define the angular momentum operator 

by J = J x e x + J y e y + J z e z which yields J n = J ■ n. Now let us find 
out what the commutation relations between the different components 
of the angular momentum operator are. We use the fact that we have 
a representation of the rotation group, i.e. Eq. (3.56) is satisfied. 
Eq. (3.55) then implies that for an infinitesimal rotation we have the 
operator equation 

e iJ x e/h e iJye/h e -ij x e/h e -ijye/h _ ^-ij z t 2 /h 

Expanding both sides to second order in e, we find 
1 + [J y , J x ]e 2 /h 2 = 1- iJ z e 2 /h 
which gives rise to the commutation relation 

[J xi Jy] = ihJz ■ (3.57) 

In the same way one can obtain the commutation relations between any 
other component of the angular momentum operator, which results in 

[Jj, Jj] = ihe ijk J k . (3.58) 

This fundamental commutation relation is perhaps the most important 
feature of the quantum mechanical angular momentum and can be used 
as its definition. 



3.3.3 Angular momentum eigenstates 

Using the basic commutation relation of angular momentum operators 
Eq. (3.58), we will now derive the structure of the eigenvectors of the 
angular momentum operators. Eq. (3.58) already shows that we are 
not able to find simultaneous eigenvectors to all the three components 
of the angular momentum. However, the square of the magnitude of 

the angular momentum J 2 = J 2 + J 2 + J% commutes with each of its 
components. In the following we will concentrate on the z-component 
of the angular momentum, for which we find 

[>, J z ] = . (3.59) 

As the two operators commute, we can find simultaneous eigenvectors 


for them, which we denote by \j,m). We know that J is a positive 
operator and for convenience we chose 

f*\j,m)=ti 2 j(j + l)\j,m) . (3.60) 

By varying j from zero to infinity the expression j(j + 1) can take any 
positive value from zero to infinity. The slightly peculiar form for the 

— * 

eigenvalues of J 2 has been chosen to make all the following expressions 
as transparent as possible. J z is not know to be positive, in fact it isn't, 
and we have the eigenvalue equation 

J z \j,m) = hm\j,m) . (3.61) 

Given one eigenvector for the angular momentum we would like to have 
a way to generate another one. This is done by the ladder operators 
for the angular momentum. They are defined as 

J± = J x ± iJ y . (3.62) 

To understand why they are called ladder operators, we first derive 
their commutation relations with J z . They are 

[4 4] = ±m ± 




Using these commutation relations, we find 

j z (j+\j,m)) = (J + J z + hJ + )\j,m) 
= (J + hm + HJ + ) \j, m) 
= h(m+l)J + \j,m) . (3.64) 

This implies that 

{J+\j,m)) = a+(J,m)\j,m + l) , (3.65) 

i.e. J + generates a new eigenvector of J z with an eigenvalue increased 
by one unit and a + (j, m) is a complex number. Likewise we find that 

(J-\j,m)) = a-(j,m)\j,m - 1) , (3.66) 

i.e. J- generates a new eigenvector of J z with an eigenvalue decreased 
by one unit and a_ (j, m) is a complex number. Now we have to es- 
tablish the limits within which m can range. Can m take any value or 
will there be bounds given by j? To answer this we need consider the 
positive operator 

> - J 2 Z = Jl + Jy . (3.67) 
Applying it to an eigenvector, we obtain 

(j,m\(fi-j 2 z )\j,m) = h 2 (j(j + l)-m 2 )>0 . (3.68) 
We conclude that the value of m is bounded by 

|m| < + . (3.69) 

While from Eqs. (3.65-3.66) we can see that the ladder operators allow 
us to increase and decrease the value of m by one unit, Eq. (3.69) 
implies that there has to be a limit to this increase and decrease. In 
fact, this implies that for the largest value m max we have 

J + \j,m max ) = 0\j,m max + 1) = , (3.70) 

while for the smallest value m min we have 

J-\j,m min ) = 0\j,m min - 1) = . (3.71) 



This implies that we have 

J-J+\j,m max ) = , (3.72) 
J+J-\j,m min ) = . (3.73) 


To determine m max and m min we need to express J_J + and J + J_ in 


terms of the operators J and J z . This can easily be done by direct 
calculation which gives 

J+J- = (Jx+iJy)(Jx-iJ y ) = J 2 +J 2 -i[J x , J y ] = J 2 -J 2 z +hJ z , (3.75) 

J-J+ = (J x -iJy)(J x +iJ y ) = J x +Jy+i[J*J y ] = f 2 -J 2 z -hJ z . (3.76) 
Using these expressions we find for m max 

= J-J+\j, m max ) = (J 2 - J 2 Z - hJ z )\j, m max ) 

= h 2 (j(j + 1) - m max (m max + l))\j,m max ) . (3.77) 

This implies that 

m max =j . (3.78) 

Likewise we find 

= J+ J_ \j, m min ) = (J 2 - J 2 Z + %J Z ) \j, m min ) 

= h 2 (j(j + 1) + m min (l - m min ))\j,m min ) . (3.79) 

This implies that 

m min = -j . (3.80) 

We have to be able to go in steps of one unit from the maximal value 
f^max to m m i n . This implies that 2j is a whole number and therefore 
we find that 


j = - with ne N . (3.81) 

Every real quantum mechanical particle can be placed in one of two 
classes. Either it has integral angular momentum, 0, 1, . . . (an example 



is the orbital angular momentum discussed in the previous section) and 
is called a boson, or it has half-integral angular momentum |, |, . . . (the 
electron spin is an example for such a particle) and is called a fermion. 
Finally we would like to determine the constant a± (j, m) in 

j±\j,m) = a±(j,m)\j,m± 1) . (3.82) 

This is easily done by calculating the norm of both sides of Eq. (3.82). 

|a+0'»"*)| 2 = ll^+b'> m )H 2 = (j, m \(J 2 ~ Jl -hJ z )\j,m) 

= h(j(j + l)-m(m + l)) . (3.83) 

Analogously we find 

|a_(j,m)| 2 = \\.L\j,m)\\ 2 = (j,m\(.P-J 2 z +hJ z )\j,m) 

= H(j(j + 1) -m(m-l)) . (3.84) 

We chose the phase of the states \j,m) in such a way that a±(j,m) is 
always positive, so that we obtain 

j+\j,™>) = hyjj(j + 1) -m(m + i)\j,m+l) (3.85) 

J-\j,m) = hdj(j + 1) - m(m - l)\j,m - 1) . (3.86) 

3.4 Addition of Angular Momenta 

In the previous section I have reviewed the properties of the angu- 
lar momentum of a single particle. Often, however, you are actually 
holding a quantum system consisting of more than one particle, e.g. 
a hydrogen atom, and you may face a situation where more than one 
of those particles possesses angular momentum. Even a single particle 
may have more than one angular momentum. An electron in a central 
potential, for example, has an orbital angular momentum as well as 
an internal angular momentum, namely the spin. Why do we need to 
add these angular momenta? To see this, consider the example of an 
electron in a central potential. 



The Hamilton operator of a particle in a central potential is given 


H = £- + V(\2\) , (3.87) 

where p is the linear momentum of the particle, m its mass and V(|x|) 
is the operator describing the central potential. From subsection 3.2.3 
we know that under this Hamilton operator the orbital angular mo- 
mentum is preserved because the Hamilton operator is invariant under 
rotations around an arbitrary axis. However, if you do experiments, 
then you will quickly realize that the Hamilton operator Hq is only the 
first approximation to the correct Hamilton operator of an electron in 
a central potential (you may have heard about that in the 2nd year 
atomic physics course). The reason is, that the electron possesses an 
internal angular momentum, the spin, and therefore there have to be 
additional terms in the Hamilton operator. These additional terms can 
be derived from a relativistic theory of the electron (Dirac equation), 
but here I will just give a heuristic argument for one of them because 
the full description of the Dirac equation would take far too long.) Intu- 
itively, an electron in a central potential rotates around the origin and 
therefore creates a circular current. Such a circular current gives rise to 
a magnetic field. The spin of an electron on the other hand gives rise to 
a magnetic moment of the electron whose orientation depends on the 
orientation of the spin - which is either up or down. This means that 
the electron, has different energies in a magnetic field B depending on 

the orientation of its spin. This energy is given by —p,oSB, where —f^oS 

is the magnetic moment of the electron and S = h(cr + <7 2 e 2 + <J 3 e 3 ) is 
the electron spin operator. The magnetic field created by the rotating 
electron is proportional to the orbital angular momentum (the higher 
the angular momentum, the higher the current induced by the rotating 
electron and therefore the higher the magnetic field) so that we find 
that the additional part in the Hamilton operator Eq. (3.87) is given 

H 1 = -ZS®Z = t(S 1 ®L 1 + S2®iv + S 3 ®L 3 ) , (3.88) 

where £ is a constant. Note that the operators S and L act on two 



— * 

different Hilbert spaces, S acts on the 2-dimensional Hilbert space of 

— * 

the electron spin and L on the Hilbert space describing the motion of 
the electron. 

Under the Hamilton operator Eq. (3.87) the spin was a constant 

of motion as the spin operator S evidently commutes with the Hamil- 
tonian Eq. (3.87), i.e. [H ,Si] = 0. For the total Hamilton operator 

H = H + Hi, however, neither the orbital angular momentum L nor 

— * 

the spin angular momentum S are constants of motion anymore. This 
can easily be seen by checking that now the commutators [Li, H] and 
[Si, H] are non- vanishing! See for example that 

[Li,H] = [Li,Hi] 

= -{[L^St-Li] 


= -ih£S 2 L 3 + ihZS 3 L 2 (3.89) 


[Si,H] = [Si, Hi] 

- -{[Si^St-Li] 


= -£,ihS 3 L 2 + t,ihS 2 L 3 

which both are evidently non-zero. However, the sum of the two spin 

operators J = L + S is a conserved quantity because all its components 
commute with the Hamiltonian, i.e. 

[Li + Si,H] = . (3.90) 

You may check this easily for the first component of Ji using the two 
commutators Eqs. (3.89-3.90). 

We know a basis of eigenstates to the orbital angular momentum 
and also one for the spin angular momentum. However, as the an- 
gular momenta separately are not conserved anymore such a choice 



of eigenstates is quite inconvenient because eigenstates to the angular 

momenta L and S are not eigenstates of the total Hamilton operator 
H — Hq + H\ anymore. Of course it would be much more convenient 
to find a basis that is composed of simultaneous eigenvectors 
of the total angular momentum and the Hamilton operator. 

This is the aim of the following subsection which describes a general 
procedure how one can construct the eigenvectors of the total angu- 
lar momentum from the eigenstates of orbital angular momentum and 
spin. As different components of the total angular momentum do not 
commute, we can of course only find joint eigenstates of the total angu- 

lar momentum operator J , its z-component J z and the total Hamilton 
operator H + Hi. 

3.4.1 Two angular momenta 

First I will present the general idea behind the addition of two angular 

momenta represented by the operators j for the first particle and j 
for the second particle. Then I will give the simplest possible explicit 

example. Note that I am writing j instead of the more precise j <E> i 
to shorten the equations a little bit. 

Now I would like to consider the total angular momentum 

% ^(2) 

J:=j +3 , (3-91) 

Here the upper bracketed index indicates on which particle the operator 
is acting. 

Like in the previous section J commutes with each of its compo- 

nents. (Check e.g. that [J ,J Z ] = [J + j| 2) ] = 0). Therefore I 

would like to find a set of joint eigenvectors of J and the z-component 
of the total angular momentum J z . These eigenvectors which are writ- 
ten as | J, M) have to satisfy 

- 2 

J \J,M) = hJ{J + l)\J,M) 
J Z \J,M) = hM\J,M) . 



As the eigenstates m^; j( 2 \ mP^) of the separate angular momenta 
and form a basis, we have to be able to write | J, M) as linear 
combinations of the eigenvectors to the individual angular momenta 
\j^\m^;j^\m^) of which there are (2jW + l)(2j( 2 ) + 1) states. It 
might seem difficult to find these linear combinations, but luckily there 
is a general recipe which I am going to explain in the following. 

In the previous section we constructed all the possible angular mo- 
mentum eigenstates of a single particle by starting with the state with 
highest m-quantum number and then working our way down by apply- 
ing the angular momentum ladder operator j_ = j x — ij y - We will apply 

— * 

an analogous strategy for the total angular momentum J. Firstly we 
need to identify the ladder operators for the total angular momentum. 
They are 

J- = J a) +J'i 2) (3.92) 
4 = jf+Jf • (3-93) 

Let us check whether these operators satisfy commutation relations 
analogous to Eq. (3.63). 

[44] = [F+F^+F] 

= ±hj?±hW 

= ±hJ± . (3.94) 

The commutation relation Eq. (3.94) allows us, just as in the case of a 
single angular momentum, to obtain the state \J,M — 1) from the state 
\J,M) via 

L\J,M) = h^J(J+l) - M(M - 1)| J, M - 1) . (3.95) 

This relation can be derived in exactly the same way as relation Eq. 
(3.86). The procedure for generating all the states \J,M) has three 
steps and starts with 

Step 1: Identify the state with maximal total angular momentum and 
maximal value for the M-quantum number. If we have two angular 



momenta and then their sum can not exceed J ma x = + ■ 

Which state could have total angular momentum J max = and 

the M = J max 7 Clearly this must be a combination of two angular 
momenta that are both parallel, i.e. we guess they should be in the 


\J = J W + J (2) , M = jW + = jW) ® f)) . (3.96) 
To check whether this assertion is true we need to verify that 

J \ J,M) = h 2 J(J+ 1)1 J, M) 
J Z \J,M) = UM\J,M} , 

with J — M — I will not present this proof here, but will let 

you do it in the problem sheets. Now follows 

Step 2: Apply the ladder operator defined in Eq. (3.92) using Eq. 
(3.95) to obtain all the states \J, M). Repeat this step until you reached 
state \J,M— - J). 

However, these are only 2J max + 1 states which is less than the to- 
tal of (2ji + l)(2j 2 + 1) states of the joint state space of both angular 
momenta. To obtain the other states we apply 

Step 3: Identify the state that is of the form | J — 1, M — J — 1) and 
then go to step 2. The procedure stops when we have arrived at the 
smallest possible value of J, which is obtained when the two angular 
momenta are oriented in opposite direction so that we need to subtract 
them. This implies that J m in = \j^ The absolute value needs to 

be taken, as the angular momentum is by definition a positive quantity. 

Following this strategy we obtain a new basis of states consisting of 

Ti=r mm {2i + \) = Ef=<r 2i + 1 - e/=cT 1 2* + 1 = ( J max + 1) 2 - J 2 mm = 

+ J (2) + l) 2 - - 3 {2) f = (2j (1) + l)(2j( 2 ) + 1) states. 
As this is a rather formal description I will now give the simplest 
possible example for the addition of two angular momenta. 


Now I am going to illustrate this general idea by solving the explicit 
example of adding two spin-|. 



The angular momentum operators for a spin-| particle are 

j x = hcr x j y = ha y j z = M z (3.97) 

with the Pauli spin-operators <7j. It is easy to check that this definition 
satisfies the commutation relations for an angular momentum. The an- 
gular momentum operator of an individual particle is therefore given by 

j = h (a x e x + o y ty + a z e z ) and the total angular momentum operator 
is given by 

J = j +j . (3.98) 
The angular momentum ladder operators are 

j- = :i m +3 m 

J + = iS'+if ■ 

The maximal value for the total angular momentum is J = 1. The 
state corresponding to this value is given by 

|J = 1,M=1) = \jV = ±, m n = ±)®\jW = ±, m W = ±) 


or using the shorthand 

IT) = = = \) (3.100) 

I I) = \3 {1) = \M l) = -\) (3.101) 

we have 

|J=1,M = 1) = |T)®|T) . 

Now we want to find the representation of the state \J = 1,M = 0) 
using the operator J_. We find 

| J=1 ,M = 0> = J-U=l^ = l> 

1 1 hV2 

G w +3i 2) )l T) ^> I T) 

- t> + m®u> ,3.102) 



To find the state \ J — 1, M — — 1) we apply J_ again and we find that 
|J= !,*=-!) = -U^LM-P) 

(ji 1) +ji 2) )|J = l,M = 0) 

= I I)® I I) (3.103) 

Now we have almost finished our construction of the eigenstates of the 
total angular momentum operator. What is still missing is the state 
with total angular momentum J = 0. Because J = this state must 
also have M = O.The state \J = 0,M = 0) must be orthogonal to 
the three states |J = 1,M = 1),|J = 1, M = 0), | J = 1, M = -1). 
Therefore the state must have the form 

\J = 0,M = 0)= li) ^ T) - IT) ^' • (3.104) 

To check that this is the correct state, just verify that it is orthogonal 
to | J = 1, M = 1), | J = 1, M = 0), | J = 1, M = -1). This concludes 
the construction of the eigenvectors of the total angular momentum. 

3.5 Local Gauge symmetries and Electro- 


Chapter 4 

Approximation Methods 

Most problems in quantum mechanics (like in all areas of physics) 
cannot be solved analytically and we have to resort to approximation 
methods. In this chapter I will introduce you to some of these meth- 
ods. However, there are a large number of approximation methods in 
physics and I can only present very few of them in this lecture. 

Often a particular problem will be in a form that it is exactly solv- 
able would it not be for a small additional term (a perturbation) in the 
Hamilton operator. This perturbation may be both, time-independent 
(e.g. a static electric field) or time-dependent (a laser shining on an 
atom). It is our task to compute the effect that such a small perturba- 
tion has. Historically these methods originated from classical mechanics 
when in the 18th century physicists such as Lagrange tried to calculate 
the mechanical properties of the solar system. In particular he and his 
colleagues have been very interested in the stability of the earths orbit 
around the sun. This is a prime example of a problem where we can 
solve part of the dynamics exactly, the path of a single planet around 
the sun, while we are unable to solve this problem exactly once we take 
into account the small gravitational effects due to the presence of all 
the other planets. There are quite a few such problems in quantum 
mechanics. For example we are able to solve the Schrodinger equation 
for the hydrogen atom, but if we place two hydrogen atoms maybe 
100 Angstrom away from each other then the problem has no exact 
solution anymore. The atoms will start to perturb each other and a 
weak perturbation to their dynamics will be the effect. While not be- 




ing strong enough to form a molecule this effect leads to an attractive 
force, the van der Waals force. This force, although being quite weak, 
is responsible for the sometimes remarkable stability of foams. 

4.1 Time-independent Perturbation The- 

In the second year quantum mechanics course you have seen a method 
for dealing with time independent perturbation problems. I will red- 
erive these results in a more general, shorter and elegant form. Then I 
will apply them to derive the van der Waals force between two neutral 

4.1.1 Non-degenerate perturbation theory 

Imagine that you have a system with a Hamilton operator H , eigen- 
vectors \4>i) and corresponding energies t; L . We will now assume that 
we can solve this problem, i.e. we know the expressions for the eigen- 
states \<pi) and energies q already. Now imagine a small perturbation 
in the form of the operator XV is added to the Hamilton operator. The 
real parameter A can be used to count the order to which we perform 
perturbation theory. Now we have the new total Hamilton operator 
H = H + XV with the new eigenvectors \ipi(X)) and energies Ei(X). 
For A = we recover the old unperturbed system. For the new system, 
the time-independent Schrodinger equation now reads 

(H + \VM(X))=E i (X)\ip i (X)) . (4.1) 

For the following it is useful to introduce the slightly unusual 'normal- 
ization ' 

(<t>i\A{X)) = 1 (4.2) 

for the perturbed eigenstates |^(A)), i.e. we do not assume that 
(^(A)hMA)) = l! 

Now we can multiply the Schrodinger equation (4.1) from the left 
with and find 

(&|(if„ + XV)MX)) = (0 i | J E7 i (A)|Vi(A)> . (4.3) 


Using the 'normalization' relation Eq. (4.2) we then obtain 

E i (\)=e i + \{<l> i \V\1> i (\)) . (4.4) 

To obtain the energy eigenvalues of the perturbed Hamilton operator 
we therefore need to compute the corresponding eigenvector ^(A)). In 
the following I will do this for the state \<f> ) to obtain the perturbed 
energy E (X). (You can easily generalize this to an arbitrary eigenvalue, 
just by replacing in all expressions O's by the corresponding value.) For 
the moment I will assume that the eigenvector \<p ) of the unperturbed 
Hamilton operator Hq is non-degenerate. You will see soon why I 
had to make this assumption. 

In order to do the perturbation theory as systematic as possible I 
introduce the two projection operators 

Po = I0o)(0o| and Q = l-P • (4.5) 

The second projector projects onto the subspace that supports the cor- 
rections to the unperturbed wave-function, i.e. 

QolV'o(A)) = |Vo(A)) - |0 O ) (4.6) 

Now let us rewrite the Schrodinger equation introducing an energy e 
which will be specified later. We write 

(e - E (X) + AV)|Vo(A)> = (e - # )|^ (A)> ■ (4-7) 
This can be written as 

|Vo(A)> = (e - Ho)- 1 ^ - E (X) + \V)\MV) ■ ( 4 -§) 
Multiplying this with the projector Q using Eq. (4.6) we find 

|^o(A)) = |0 O ) + Qo(e - H )-\e - E (X) + AT>)|Vo(A)> . (4.9) 
Now we can iterate this equation and we find 

l^o(A)) = £ [Qo(e - H )-\e - E (X) + \V)] n |0 O ) . (4.10) 




Now we can plug this into Eq. (4.4) and we find the expression for the 
perturbed energies 


#o(A) = e + E(^o|AV [Q (e - H y\e - E (X) + AV")]" |0 O ) • 



Eqs. (4.9) and (4.11) give the perturbation expansion of the energies 
and corresponding eigenvectors to all orders. 

Remark: The choice of the constant e is not completely arbitrary. 
If it is equal to the energy of one of the excited states of the unperturbed 
Hamilton operator, then the perturbation expansion will not converge 
as it will contain infinite terms. This still leaves a lot of freedom. 
However, we also would like to be able to say that by taking into account 
the first k terms of the sum Eq. (4.11), we have all the terms to order 
X k . This is ensured most clearly for two choices for the value of e. One 
such choice is evidently e = E (X). In that case we have 


E (X) = e + ]>>o|AV [Qo(E (\) - H y\\V)] n |0 O ) . (4.12) 


This choice is called the Brillouin-Wigner perturbation theory. The 
other choice, which is often easier to handle, is that where e = e . In 
that case we obtain 


£o(A) = e + E(^o|AV [<? (eo - H )-\e - E (X) + XV)]" \<f> ) ■ 



This is the Rayleigh perturbation theory. For this choice (eo — Eq(X) + 
XV) is proportional to A and therefore we obtain higher order correc- 
tions in A by choosing more terms in the sum. 

Now let us look at the first and second order expression so that you 
can compare it with the results that you learned in the second year 
course. To do this we now specify the value of the arbitrary constant 
e to be the unperturbed energy e . For the energy of the perturbed 
Hamilton operator we then obtain 

E (X) = e + (0o|AV|0o) + (<Po\WQ (e - H )-\e - E Q (X) + XV)\<f) ) + 


= e + A(0o|%o> + (0 o |AV £ 10^X^1 — ^ — (e - E (\) + \V)\<f> ) + 

= e + A((/>o|^|0o) + A 2^ + ••• ( 4 - 

Remark: If |0 O ) is the ground state of the unperturbed Hamilton oper- 
ator, then the second order contribution in the perturbation expansion 
is always negative. This has a good reason. The ground state of the 
total Hamilton operator H = H + V is j^o) and we find for |0 O ) that 

E = (^ \H + V\rl> ) < (MHo + %o> = E - £ 

This implies that the second order perturbation theory term is negative! 

The new eigenstate to the eigenvector Eq. (4.14) can be calculated too, 
and the first contributions are given by 

l^o) = \<P ) + Q (e-H )- 1 (e -E + V)\<f ) o) + ... 

= 100) + E l0n)(0n|^^(e O - E + V)\(f> ) + ... 
= |0o) + £|0n)^^(0n|Vl0o) + --- ■ 

Looking at these two expressions it is clear why I had to demand, 
that the eigenvalue e of the unperturbed Hamilton operator H has 
to be non-degenerate. In order to obtain a meaningful perturbation 
expansion the individual terms should be finite and therefore the factors 
— — need to be finite for n ^ 0. 

4.1.2 Degenerate perturbation theory 

This part has not been presented in the lecture 

What do we do in the case of degenerate eigenvalues? We can follow 
a similar strategy but as you can expect things get a little bit more 



Let us assume that we have a Hamilton operator Hq, eigenvectors 
\4>\) and corresponding energies e« where the upper index v numerates 
an orthogonal set of eigenvectors to the degenerate eigenvalue Again 
let us deal with the eigenvalue e . 

Now we write down the projectors 

Po = £l0o>(0ol and Q = l-P . (4.15) 

Then we find 

Qo|<(A)> = |<(A)> - Po|<(A)> , (4.16) 

where the eigenvector I^q) of the perturbed Hamilton operator origi- 
nates from the unperturbed eigenvector 

To obtain the analogue of Eq. (4.3) we multiply the Schrodinger 
equation for the state ^(A)) from the left with (^"(A)|Po an d we 

(^(A)|P (P + - Eg(\))\r(\)) = . (4.17) 

Note that I have proceeded slightly different from the non-degenerate 
case because there I multiplied from the left by {<j>"\. I could have done 
that too, but then I would have been forced to introduce a 'normal- 
ization' condition of the form (^-^(A)) = 5^ u which is quite hard to 

Eq. (4.17) will eventually help us determine the new energy eigen- 
values. However, this time this will involve the diagonalization of a 
matrix. The reason of course being that we have a degeneracy in the 
original unperturbed eigenvalue. 

Now we proceed very much along the lines of the unperturbed per- 
turbation theory. Using Eq. (4.16) we rewrite Eq. (4.9) and obtain 

|^(A))=P |^(A))+Qo(eo-Po)- 1 (eo-P M (A)+AT>)|^(A)) . (4.18) 
Iterating this equation yields 


IC(A)) = E[Qo( e o-Po)- 1 ( e -Po M (A) + AV)] n p |<(A)) . (4.19) 



This implies that we can calculate the whole eigenvector from 
the eigenvalues E fl (X) and the components Po\ipo(X)) of the eigenvector 
in the eigenspace of the eigenvalue eo of the Hamilton operator H . 

This may now be inserted in Eq. (4.17) so that we find with the 
shorthand notation \ipQ P ) = -Pol^o) 


W P |(^o+AV-^(A)) £ [Q (e - H Q )-\e - Eg(X) + \V)] n \^ p ) = 



Now we need to determine the energies Eft from Eq. (4.20). Let us 
first determine the lowest order approximation, which means that we 
set n — 0. We find for any /j, and v that 

(^ P |e + XV - E»(X)\W P ) = . (4.21) 

Therefore the new energy eigenvalues are obtained by diagonalizing 
the operator V in the eigenspace of the eigenvalue eo of the Hamilton 
operator H spanned by the \4>q). We find 

Eq(X) = e + (4.22) 

where the V w are the eigenvalues of V. 

The second order contribution is then obtained by using the eigen- 
vectors obtained from Eq. (4.21) and use them in the next order expres- 
sion, e.g. for n = 1. The second order expression for the perturbation 
theory is therefore obtained from taking the expectation values 

(^' P |(e + W w - Eft(X)) + \VQ (e - H^XV^) = . 
This gives 

^, + At + A'E^^. (4.23) 
mf L e e m 

In the following section I will work out an example for the use of per- 
turbation theory which gives a non-trivial result. 



4.1.3 The van der Waals force 

In this section I am going to derive the force that two neutral atoms 
are exerting on each other. This force, the van der Waals force, can 
play a significant role in the physics (also the chemistry and mechanics) 
of neutral systems. In the derivation of the van der Waals force I will 
use time-independent perturbation theory. The calculations will be 
simplified by the fact that the system of two atoms exhibits symmetries 
which have the effect, that the first order contribution to perturbation 
theory vanishes. 

The basic experimental situation is presented in Fig. 4.1. We dis- 
cuss the van der Waals force between two hydrogen atoms. The two 

Figure 4.1: Two hydrogen atoms are placed near each other, with a 
distance R much larger than the Bohr radius ao- 

atomic nuclei (proton) are positioned at a fixed distance from each 
other. We assume that the protons are not moving. The proton of 
atom A is placed at the origin, while the proton of atom B is located at 

— * 

the position R. The electron of atom A is displaced from the origin by 
while atom B is displaced from the nucleus B by r#. To make sure 
that we can still speak of separate hydrogen atoms we need to demand 
that the separation between the two protons is much larger than the 
radius of each of the atoms, i.e. 

\R\ > a , (4.24) 

where ao is the Bohr radius. 

After we have described the physical situation we will now have to 
determine the Hamilton operator of the total system. The Hamilton 


operator has the form 

H = H + V , (4.25) 

where V describes the small perturbation that gives rise to the van der 
Waals force. The unperturbed Hamilton-operator H is that of two 
hydrogen atoms, i.e. 

H = — + — K L- , (4.26) 

2m 2m A7re \f A \ ^e \f B \ 

which can be solved exactly. We have energy eigenfunctions of the 
form |0^ im )|0n',z',m') with energies E n + E n >. The perturbation to this 
Hamilton operator is due to the electrostatic forces between electrons 
and protons from different atoms. From Fig. 4.1 we can easily see that 
this gives 

e 2 e 2 e 2 e 2 

V = + =; ; ; =; ; ^ ^~ . 

4ne R 47re |-R + f B - f A \ 4ire \R + r B \ 47re |-R + f A | 


As we have assumed that the Bohr radius is much smaller than the 
separation between the two hydrogen atoms, we can conclude that 
\ta I, \tb\ ^ R- Therefore we are able to expand the perturbation 
Hamilton operator and we obtain after lengthy but not too difficult 

V dd = 

e 2 


r A r B _ Ar A R u ){f B R u ) 
IP ° ' IP 


where R u is the unit vector in direction of R. Now we are in a position to 
determine the first and second order contribution of the perturbation 
theory. Because we are dealing with the ground state, we have no 
degeneracy and we can apply non-degenerate perturbation theory. 

First order correction The first order contribution to perturbation 
theory is given by 

AE 1 = « ,ol«o,olV r *iWo,o>Wo,o> > ( 4 - 29 ) 



where \4>oofi) are the unperturbed ground-states of atom A/B. Inserting 
Eq. (4.27) into Eq. (4.29) gives 


(#?o,o \fA |0^o,q) (0^o,o \r B \(j)o fifi ) 
i? 3 


m,o,o\r aRu I 06%) {n fifi \r B Ru \ <j>o fifi ) 
R 3 


Now we can use a symmetry argument to show that this whole, rather 
lengthy expression must be zero. Why is that so? What appears in 
Eq. (4.30) are expectation values of components of the position op- 
erator in the unperturbed ground state of the atom. However, the 
unperturbed Hamilton operator of the hydrogen atom possesses some 
symmetries. The relevant symmetry here is that of the parity, i.e. the 
Hamilton operator Hq commutes with the operator P which is defined 
by P\x) = | — x). This implies that both operators, H and P, can 
be diagonalized simultaneously, i.e. they have the same eigenvectors. 
This is the reason why all eigenvectors of the unperturbed Hamilton 
operator Hq are also eigenvectors of the parity operator. Therefore we 
have P|#Jm)) = ±l#u),o)- This implies that 

(00,0,0 \A 0O,O,o) 

Vo,o| - z|0o,o,o) 

This equality implies that (0o,o,o|^|0o,o,o) = and therefore AE X = 0! 
This quick argument illustrates the usefulness of symmetry arguments. 

Second order correction The second order contribution of the per- 
turbation theory is given by 


(n' ,1' ,m'),(n,l,m) 

^^^,wl(0n, ^ , m l^l06 4 o,o)l<o,o)P 

2En — E„i — E n 


where the J2' means that the state |0o 4 oo)l0o 3 oo) * s excluded from the 
summation. The first conclusion that we can draw from expression Eq. 


(4.31) is that the second order correction AE 2 is negative so that we 

AE 2 = (4.32) 

with a positive constant C . The force exerted on one of the atoms is 
therefore given by 

dE 6C . 

and therefore the force aims to reduce the distance between the atoms 
and we conclude that the van der Waals force is attractive. This is also 
the classical result that has been known before the advent of quantum 

The physical interpretation of the van der Waals force is the fol- 
lowing: A hydrogen atom should not be viewed as a stationary charge 
distribution but rather as a randomly fluctuating electric dipole due 
to the motion of the electron. Assume that in the first atom a dipole 
moment appears. This generates an electric field, which scales like 
with distance. This electric field will now induce an electric dipole mo- 
ment in the other atom. An electric dipole will always tend to move 
towards areas with increasing field strength and therefore the second 
atom will be attracted to the first one. The induced dipole moment 
will be proportional to the electric field. Therefore the net force will 
scale as — jjijp- This semi-classical argument gives you an idea for the 
correct /^-dependence of the van der Waals force. 

However, this semi-classical explanation fails to account for some of 
the more intricate features of the van der Waals force. One example 
is that of two hydrogen atoms, one in the ground state and one in an 
excited state. In that case the van der Waals force scales as and 
the constant C can be either positive or negative, i.e. we can have a 
repulsive as well as an attractive van der Waals force. The computation 
of this case is slightly more complicated because excited states in the 
hydrogen atom are degenerate, which requires the use of degenerate 
perturbation theory. On the other hand it only involves first order 



4.1 .4 The Helium atom 

Now let us try to apply perturbation theory to the problem of the 
Helium atom. In fact it is not at all clear that we can do that because 
the two electrons in the Helium atom are very close together and their 
mutual exerted force is almost as strong as that between the nucleus 
and the individual electrons. However, we may just have a look how far 
we can push perturbation theory. Again I am interested in the ground 
state of the Helium atom. The Hamilton operator is given by 

Va Vb 2e2 2e 2 e 2 , 4n4 s 

H = 7T- + 7T~ 5 ~ H 5 ~ • ( 434 ) 

2m 2m 47reo|r\4| 47reo|rB| 4neo\rA — r B \ 

The first four terms, Hq describe two non-interacting electrons in a 
central potential and for this part we can find an exact solution, given 
by the energies and wave-functions of a He + ion. Now we perform 
perturbation theory up to first order. The zeroth order wavefunction 
is given by 

\tl>He) = \^He+) ® \^He+) (4.35) 

where \ipHe+) is the ground state of an electron in a Helium ion, i.e. a 
system consisting of two protons and one electron. In position space 
this wavefunction is given by 

^Heir) = —^e -o , (4.36) 

where ao = 0.511 10~ 10 m is the Bohr radius and Z = 2 is the number 
protons in the nucleus. The energy of this state for the unperturbed 
system is given by E = —108.86^. The first order correction to the 
energy of the ground state is now given by 


AE — [ d 3 r A j d 3 r B \^ He (r)\ 2 ^ ^- = 34eV . (4.37) 

Therefore our estimate for the ground state energy of the Helium atom 
is E 1 = -108.8eV + 34eV = -7A.8eV . This compares quite well with 
the measured value of E 1 = — 78.9eV. However, in the next section 
we will see how we can actually improve this value using a different 
approximation technique. 


4.2 Adiabatic Transformations and Geo- 
metric phases 

4.3 Variational Principle 

After this section on perturbation methods I am now moving on to a 
different way of obtaining approximate solutions to quantum mechan- 
ical problems. Previously we have investigated problems which were 
'almost' exactly solvable, i.e. the exactly solvable Hamilton operator 
has a small additional term. Now I am going to deal with problems 
which can not necessarily be cast into such a form. An example would 
be the Helium atom (which cannot be solved exactly) as compared to 
a negative Helium ion (which is basically like a hydrogen atom and 
therefore exactly solvable). Because the two electrons in the Helium 
atom are very close, it is not obvious that a perturbation expansion 
gives a good result, simply because the electrostatic forces between the 
electrons are almost as large as those between electrons and nucleus. 
Here variational principles can be very useful. The example of the He- 
lium atom already indicates that variational methods are paramount 
importance e.g. in atomic physics, but also in quantum chemistry or 
solid state physics. 

Here I will explain a simple variational method, which nevertheless ex- 
hibits the general idea of the method. Variational methods are partic- 
ularly well suited to determine the ground state energy of a quantum 
mechanical system and this is what I will present first. The whole 
method is based on the following theorem concerning the expectation 
value of the Hamilton operator. 

Theorem 49 Given a Hamilton operator H with ground state |0o) an d 
ground state energy E . For any we have 

4.3.1 The Rayleigh-Ritz Method 




Proof: For the proof we need to remember that any state vector 
can be expanded in terms of the eigenvectors \<pi) of the Hermitean 
operator H. This means that we can find coefficients such that 

|V> . (4.39) 


I will use this in the proof of theorem 49. 


= J2 a i a j E i( ( f ) i\ ( f ) j} 

= EN 2 ^ • 


Now we can see that 

^\H\^)-E = EM 2 £*-£o 


> o , 

because Ei > E . If the lowest energy eigenvalue E is not degenerate 
then we have equality exactly if = |0o) □• 

How does this help us in finding the energy of the ground state of a 
quantum mechanical system that is described by the Hamilton operator 
HI The general recipe proceeds in two steps. 

Step 1: Chose a 'trial' wave function \i^(ai, . . . , a at)) that is parametrized 
by parameters a±, . . . , ocn- This choice of the trial wave function will 
often be governed by the symmetries of the problem. 
Step 2: Minimize the expression 

E = (i/j(a 1: . . . , a N )\H\i/;(a 1: . . . , a N )) , (4.40) 

with respect to the ctj. The value E min that you obtain this way is your 
estimate of the ground state wave function. 



The Helium atom Now let us come back to the Helium atom to see 
whether we can make use of the variational principle to obtain a better 
estimate of the ground state energy. 

In our perturbation theoretical calculation we have used the wave- 
function Eq. (4.36) with Z = 2. Now let us introduce the adjustable 
parameter a and use the trial wavefunction 

^He(r) = -^-e -o , (4.41) 


where Z e ff = Z—o. This is a physically reasonable assumption because 
each electron sees effectively a reduced nuclear charge due to the shield- 
ing effect of the other electron. Now we can use this new trial wave- 
function to calculate the energy expectation value of the total Hamilton 
operator with this wavefunction. We find after some computations 

E(a) = -2R H [z 2 - jjz + h -a - <j 2 ) , (4.42) 


where Rh = 64 ™ 3 t t n 3 c is the Rydberg constant. Now we need to com- 
pute that value of a for which Eq. (4.42) assumes its minimal value. 
The value that we obtain is a = ^, independently of the value of Z. 
Inserting this into Eq. (4.42) we find 

E mm = -2R H (Z - ^) 2 . (4.43) 

For the Helium atom (Z=2) we therefore obtain 

E mm = -77AeV , (4.44) 

for the ground state energy. This is quite close to the true value of 
— 78.9eV and represents a substantial improvement compared to the 
value obtained via perturbation theory. Making a slightly more detailed 
choice for the trial wavefunction (basically using one more parameter) 
would lead to substantially improved values. This shows that the vari- 
ational method can indeed be quite useful. 

However, the next example will illustrate the shortcomings of the 
variational method and proves that this method has to be applied with 
some care. 



The harmonic oscillator The Hamilton operator for the harmonic 

oscillator is given \mu) 2 x 2 . Now, assume that we do not 

know the ground state energy and the ground state wavefunction of 
the harmonic oscillator. This Hamilton operator commutes with the 
parity operator and therefore the eigenstates should also be chosen 
as eigenfunctions of the parity operator. Therefore let us chose the 
normalized symmetric wave-function 

ha 3 / 2 1 

Ux) = \ Y— ■ ( 4 - 45 ) 

V 7r x 2 + a 

Of course I have chosen this slightly peculiar function in order to make 
the calculations as simple as possible. Omitting the detailed calcu- 
lations we find that the mean value of the energy of the harmonic 
oscillator in this state is given by 

r ( h 2 d 2 1 \ 

{rl> a \H\rl> a ) = J ^(x)f- — — + -mw 2 a; 2 J^ a (a;)da; (4.46) 

h 2 l l 9 , 

= h -mu 2 a . (4.47) 

4m a 2 

Now we need to find the value for which this expression becomes min- 
imal. We obtain ^ 

V2 mu 

and for the energy 


Therefore our estimate gives an error 

Emin — Eq _ \[2 — 1 

E ~2~ 

E min = . (4.49) 

0.2 . (4.50) 

This is not too good, but would have been a lot better for other trial 

This example shows the limitations of the variational principle. It 
very much depends on a good choice for the trial wave function. It 
should also be noted, that a good approximation to the ground state 


energy does not imply that the chosen trial wave function will give good 
results for other physical observables. This can be seen from the above 
example of the harmonic oscillator. If we compute the expectation 
value of the operator x 2 then we find 

V Z TfioJ 

which is quite close to the true value of \hu). On the other hand, the 
expectation value of the operator x 4 diverges if we calculate it with 
the trial wave function, while we obtain a finite result for the true 
ground-state wave function. 

The variational principle for excited states. The variational 
method shown here can be extended to the calculation of the ener- 
gies of excited states. The basic idea that we will be using is that 
eigenvectors to different eigenvalues of the Hamilton operator are nec- 
essarily orthogonal. If we know the ground state of a system, or at least 
a good approximation to it, then we can use the variational method. 
The only thing we need to make sure is that our trial wavefunction is 
always orthogonal to the ground state wave function or the best ap- 
proximation we have for it. If we have ensured this with the choice of 
our trial function, then we can proceed analogously to the variational 
principle for the ground state. 

4.4 Time- dependent Perturbation Theory 

In the previous sections I have explained some of the possible methods 
of stationary perturbation theory. Using these methods we are able, in 
principle, to approximate the energy eigenvalues as well as the eigen- 
vectors of the Hamilton operator of a quantum mechanical system that 
has a small perturbation. We are then able to approximate the spectral 
decomposition of the Hamilton operator, which is given by 

H = j2E t mm , (4.5i) 


where the Ei are the energy eigenvalues of the Hamilton operator and 
the the corresponding eigenvectors. Equation (4.51) is sufficient to 



compute the time evolution of any possible initial state \<p{to)) using 
the solution of the Schrodinger equation 

e -^-'«)|0) = Yle- iEi{t - toyn \A)(i>i\<f>(to)) ■ (4.52) 


Using stationary perturbation theory we are then able to obtain the 
approximate time-evolution of the system. 

However, there are reasons why this approach is not necessarily 
the best. First of all, we might have a situation where the Hamilton 
operator of the system is time-dependent. In that case the solution 
of the Schrodinger equation is generally not of the form e - 1 f H(t')dt'/h 
anymore, simply because Hamilton operators for different times do not 
necessarily commute with each other. For time-dependent Hamilton 
operators we have to proceed in a different way in order to obtain 
approximations to the time evolution. 

There are two situations in which we may encounter time-dependent 
Hamilton operators. While the full Hamilton operator of a closed sys- 
tem is always time-independent, this is not the case anymore if we 
have a system that is interacting with its environment, i.e. an open 
system. An example is an atom that is being irradiated by a laser 
beam. Clearly the electro-magnetic field of the laser is time-dependent 
and therefore we find a time-dependent Hamilton operator. A time- 
dependent Hamilton operator may also appear in another situation. If 
we have given the Hamilton operator of a quantum system which is 
composed of two parts, the solvable part H an d the perturbation V, 
both of which may be time independent, then it can be of advantage 
to go to an 'interaction' picture that is 'rotating ' with frequencies de- 
termined by the unperturbed Hamilton operator H . In this case the 
remaining Hamilton operator will be a time-dependent version of V. 

The rest of this section will be devoted to the explanation of a 
method that allows us to approximate the time evolution operator of 
a time-dependent Hamilton operator. The convergence of this method 
is very much enhanced if we go into an interaction picture which elim- 
inates the dynamics of the system due to the unperturbed Hamilton 
operator. The definition of the 'interaction' picture in a precise man- 
ner will be the subject of the first part of this section. 


4.4.1 Interaction picture 

Often the total Hamilton operator can be split into two parts H = 
H Q + V, the exactly solvable Hamilton operator H and the pertur- 
bation V. If we have a Hamilton operator, time-dependent or not, 
and we want to perform perturbation theory it will always be useful 
to make use of the exactly solvable part H of the Hamilton operator. 
In the case of time-independent perturbation theory we have used the 
eigenvalues and eigenvectors of the unperturbed Hamilton operator and 
calculated corrections to them in terms of the unperturbed eigenvec- 
tors and eigenvalues. Now we are going to do the analogous step for 
the time evolution operator. We use the known dynamics due to the 
unperturbed Hamilton operator Hq and then determine corrections to 
this time evolution. In fact, the analogy goes so far that it is possi- 
ble in principle to rederive time-independent perturbation theory as a 
consequence of time-dependent perturbation theory. I leave it to the 
slightly more ambitious student to work this out in detail after I have 
explained the ideas of time-dependent perturbation theory. 
The Schrodinger equation reads 

ihj t m)) = (Ho + V)(t)\m) , (4-53) 

with a potentially time dependent total Hamilton operator H(t). The 
solution of the Schrodinger equation can be written formally as 

mt)) = u(t,t')mt')) , (4.54) 

where the time evolution operator U (t, t') obeys the differential equa- 

ihjUitX) = H{t)U{t,t') . (4.55) 

This can easily be checked by inserting Eq. (4.54) into the Schrodinger 
equation. Now let us assume that we can solve the time-evolution 
that is generated by the unperturbed Hamilton operator Ho(t). The 
corresponding time evolution operator is given by Uo(t,t') which gives 
the solution \ipo(t)) = U (t,t')\ip (t')) of the Schrodinger equation 




What we are really interested in is the time evolution according to 
the full Hamilton operator H . As we already know the time evolution 
operator U (t,t'), the aim is now the calculation of the deviation from 
this unperturbed time evolution. Let us therefore derive a Schrodinger 
equation that describes this deviation. We obtain this Schrodinger 
equation by going over to an interaction picture with respect to the 
time to which is defined by choosing the state vector 

Mt))=U&(t,to)m)) . (4.57) 

Clearly the state in the interaction picture has been obtained by 'un- 
doing' the part of the time evolution that is due to the unperturbed 
Hamilton operator in the state \i^{t)) and therefore describes that part 
of the time evolution that is due to the perturbation V(t). Now we have 
to derive the Schrodinger equation for the interaction picture wavefunc- 
tion and in particular we have to derive the Hamilton-operator in the 
interaction picture. This is achieved by inserting Eq. (4.57) into the 
Schrodinger equation Eq. (4.53). We find 

ihj t (Uo(t,t )\Mt))) = (H + V)(t)U (t,t )\Mt)) (4-58) 
and then 

ih (^#o(Mo)) \Mt)} +ihu (t,t Q )j t \Mt)) = (H +v)(t)Uo(t,t )\Mt)} 


Now we use the differential equation 

ih^U (t,t ) = H (t)U (t,t ) . (4.60) 

Inserting this into Eq. (4.59) gives 

H Uo(t,to)\Mt))+MU (t,t )j t \Mt)) = (Ho + V)(t)U (t,t )\Mt)) 

and finally 

ihf f \Mt)) = ul(t,t )v(t)u (t,t )\Mt)) 


Using the definition for the Hamilton operator in the interaction picture 
H I (t) = U t (t,t )(H-H )U (t,to) (4.61) 
we find the interaction picture Schrodinger equation 

ih^lMt)) = Hi(t)\Mt)) • (4-62) 

The formal solution to the Schrodinger equation in the interaction pic- 
ture can be written as 

\Mt)) = Ui(t, f)llMO> • (4-63) 

From Eq. (4.63) we can then obtain the solution of the Schrodinger 
equation Eq. (4.53) as 

\m) = u (t,to)\Mt)) 

= #o(Mo)#/(*,OI^(*')> 

= U (t,t )Ui(t, t')tf {t',to)m')) . (4.64) 

This result shows that even in a system with a time independent 
Hamilton operator H we may obtain a time dependent Hamilton opera- 
tor by going over to an interaction picture. As time dependent Hamilton 
operators are actually quite common place it is important that we find 
out how the time evolution for a time-dependent Hamilton operator 
can be approximated. 

4.4.2 Dyson Series 

Given the Schrodinger equation for a time dependent Hamilton oper- 
ator we need to find a systematic way of obtaining approximations to 
the time evolution operator. This is achieved by time-dependent per- 
turbation theory which is based on the Dyson series. Here I will deal 
with the interaction picture Schrodinger equation Eq. (4.62). To find 
the Dyson series, we first integrate the Schrodinger equation Eq. (4.62) 
formally. This gives 

\Mt)) = \Mt')) -^fidhH^lMh)) • (4-65) 



This equation can now be iterated. We then find the Dyson series 

\Mt)) = \i>i{t'))-l f dhHjfaMtf))-^ /V f L dt 2 H I {t 1 )H I {t 2 )\i, I {t l ))+... 
ii Jt' a Jf Jt' 


From this we immediately observe that the time evolution operator in 
the interaction picture is given by 

U I (t,t') = l-l f fdh [ tl dt 2 V I (t 1 )V I (t 2 ) + ... . 

a jf h Jf Jf 


Equations (4.66) and (4.67) are the basis of time-dependent perturba- 
tion theory. 

Remark: It should be pointed out that these expressions have some 
limitations. Quite obviously every integral in the series will generally 
have the tendency to grow with increasing time differences \t — t'\. For 
sufficiently large \t — t'\ the Dyson series will then fail to converge. 
This problem can be circumvented by first splitting the time evolution 
operator into sufficiently small pieces, i.e. 

U T (t, 0) = Ufa t„)#j(*n, i n _x) . . . Ujih, 0) , (4.68) 

such that each time interval U — is very small. Then each of the 
time evolution operators is calculated using perturbation theory and 
finally they are multiplied together. 

4.4.3 Transition probabilities 

For the following let us assume that the unperturbed Hamilton operator 
is time independent and that we take an interaction picture with respect 
to the time t — 0. In this case we can write 

U (t,0) = e - iAot/h . (4.69) 

Under the time evolution due to the unperturbed Hamilton operator 
the eigenstates \<f> n ) to the energy E n of that Hamilton operator 
only obtain phase factors in the course of the time evolution, i.e. 

e- l ^ t/h \<p n ) = e- iE ^ h \(t> n ) = e-^|0„) , (4.70) 


where uo n = E n /K. Under the time evolution due to the total Hamilton 
operator H this is not the case anymore. Now, the time evolution will 
generally take an eigenstate of the unperturbed Hamilton operator H 
to a superposition of eigenstates of the unperturbed Hamilton operator 
H , i-e. 

e-^-^Vn) =E«"~*(*)I&> ( 4 - 71 ) 


with nonzero coefficients a n ^k{t)- If we make a measurement at the 
later time t we will have a non-zero probability |a„^| 2 to find eigen- 
states \4> m ) with m 7^ n. We then say that the system has a transition 
probability p n ^, k (t) for going from state \4> n ) to state \4>k)- These 
transitions are induced by the perturbation V, which may for example 
be due to a laser field. Let us calculate the transition probability in 
lowest order in the perturbation V. This is a valid approximation as 
long as the total effect of the perturbation is small, i.e. the probability 
for finding the system in the original state is close to 1. To obtain 
the best convergence of the perturbative expansion, we are going over 
to the interaction picture with respect to the unperturbed Hamilton 
operator H Q and then break off the Dyson series after the term linear 
in jfj(i) = eiHoh/hy^^-iHoti/n^ If the initial state at time t > is ^ 

then the probability amplitude for finding the system in state \4> m ) at 
the later time t is given by 

a n ^ m (t) = {<p m \4>{t)) 

« (<P m \ (|0n> - \ f v dhe^^V^e-^^l^ 

= S^-^dhe^-^ifalVihMn) . (4.72) 
If m 7^ n, we find 

fln^Ww-^^^ie^-^* 1 ^!^!)!^) , (4.73) 
and the transition probability is then given by 

Pn^ m (t) = K^m(t)\ 2 . (4.74) 



Periodic perturbation 

Let us consider the special case in which we have a periodic perturbation 

V{t) = V cos out =^V (e iut + e - *"*) . (4.75) 

Inserting this into Eq. (4.73) we find using uj n 
i f f , .-. . * 1 


Static perturbation For a time independent perturbation we have 
tjj — 0. This leads to 

a n ^ m (0 = - = (<f> m \V \(f> n ) — (4.77) 

tl y iui mn J 

and then to 

Pn^W = F l(0 m |^|0n)| 2 ^^ (4.78) 

For sufficiently large times t this is a very sharply peaked function in 
the frequency oj mn . In fact in the limit t — > oo this function tends 
towards a delta-function 

&-^r = 2 ^M ■ ( 4 - 79 ) 

For sufficiently large times (i ^> u;"^) we therefore find that the tran- 
sition probability grows linearly in time. We find Fermi's golden rule 
for time independent perturbations 


vlZlti) = ttf\(<P m \V,\<P n )\H(u n - u m ) . (4.80) 

Obviously this cannot be correct for arbitrarily large times t, because 
the transition probabilities are bounded by unity. 


High frequency perturbation If the frequency of the perturbation 
is unequal to zero, we find two contributions to the transition ampli- 
tude, one with a denominator uj — uj mn and the other with the denom- 
inator uj + uj mn . As the frequency uj is always positive, only the first 
denominator can become zero, in which case we have a resonance and 
the first term in Eq. (4.76) dominates. We find 

^ / G j n 2 sin 2 ( U}+Umn ) t \ 

Pn-m(t) = ^|(0 m |Wn)| 2 1 + 1 ■ 

Again we find that for large times the transition probability grows lin- 
early in time which is formulated as Fermi's golden rule for time de- 
pendent perturbations 

Pnt° m (t) = t^\((f) m \Vo\(f) n )\\5(u n -u m -u)+5(u n -u m +oj)) . (4.82) 
The Zeno effect 

Fermi's golden rule is not only limited to times that are not too large, 
but also finds its limitations for small times. In fact for times t <C uj~^ l 
the transition probability grows quadratically in time. This is not only 
a feature of our particular calculation, but is a general feature of the 
quantum mechanical time evolution that is governed by the Schrodinger 
equation. This can be shown quite easily and is summarized in the 

Theorem 50 Given a Hamilton operator H and an initial state |0(O)) 7 
then the transition probability to any orthogonal state grows quadrat- 
ically for small times, i. e. 

lim||(0j0(t))| 2 = O . (4.83) 

Proof: Let us first calculate the derivative of the transition probability 
pit) = |a(t)| 2 with a{t) = (<p±\(p(t)) for arbitrary times t. We find 



Obviously a(t = 0) = (0j_|0(O)) = 0, so that we find 


This finishes the proof □. 

The transition probability for short times is therefore 

t 2 Ct 2 
p(t) = p(0) + —p"(0) + . . . = — — h higher orders in t 


where C is a positive constant. 

Theorem 50 has a weird consequence which you have encountered 
(in disguise) in the first problem sheet. Assume that we have a two 
state system with the orthonormal basis states |0) and |1). Imagine 
that we start at time t — in the state |0). The time evolution of the 
system will be governed by a Hamilton operator H and after some time 
T the system will be in the state |1). 

Now let the system evolve for a time T, but, as I am very curious 
how things are going I decide that I will look in which state the system 
is after times ,T. The time evolution takes the initial state 

after a time - into 

Now I make a measurement to determine in which of the two states |0) 
or |1) the system is. The probability for finding the system in state |0) 
is pi = 1 — with some nonzero constant C. If I find the system 
in the state |0), then we wait until time — and perform the same 
measurement again. The probability that in all of these measurements 
I will find the system in the state |0) is given by 

In the limit of infinitely many measurements, this probability tends to 
1. This result can be summarized as 


e-^|0) . 



A continuously observed system does not evolve in time. 


This phenomenon has the name 'Quantum Zeno effect' and has indeed 
been observed in experiments about 10 years ago. 

With this slightly weird effect I finish this part of the lecture and 
now move on to explain some features of quantum information theory, 
a field that has developed in the last few years only. 


Part II 

Quantum Information 


Chapter 5 

Quantum Information 

In 1948 Claude Shannon formalised the notion of information and cre- 
ated what is now known as classical information theory. Until about 
five years ago this field was actually known simply as information the- 
ory but now the additional word 'classical' has become necessary. The 
reason for this is the realization that there is a quantum version of the 
theory which differs in quite a few aspects from the classical version. 
In these last lectures of this course I intend to give you a flavour of this 
new theory that is currently emerging. What I am going to show you 
in the next few lectures is more or less at the cutting edge of physics, 
and many of the ideas that I am going to present are not older than 5 
years. In fact the material is modern enough that many of your physics 
professors will actually not really know these things very well. So, af- 
ter these lectures you will know more than some professors of physics, 
and thats not easy to achieve as a third year student. Now you might 
be worried that we will require plenty of difficult mathematics, as you 
would expect this for a lecture that presents the cutting edge of physi- 
cal science. This however, is not the case. In this lecture I taught you 
many of the essential techniques that are necessary for you to under- 
stand quantum information theory. This is again quite different 
from other branches of physics that are at the cutting edge of research, 
e.g. super string theory where you need to study for quite a while until 
you reach a level that allows you to carry out research. The interesting 



point about quantum information theory is the fact that it unifies two 
apparently different fields of science, namely physics and information 

Your question will now be: 'What is the connection'? and 'Why 
did people invent the whole thing' ? 

The motivation for scientists to begin to think about quantum infor- 
mation came from the rapid development of microprocessor technology. 
As many of you know, computers are becoming more and more pow- 
erful every year, and if you have bought a computer 5 years ago then 
it will be hopelessly inferior to the latest model that is on the market. 
Even more amazing is the fact that computers do get more powerful, 
but they do not get much more expensive. This qualitative descrip- 
tion of the development of micro-electronics can actually be quantified 
quite well. Many years ago, in the 1960's, Gordon Moore, one of the 
founders of Intel observed that the number of transistors per unit area 
©^ffiicVi: cffik e dBufflfeFr6^gWl^ s imMy?^^hPffi8$B& 
fk^WmftpWft w^lTOMnue in the future. As it turns out, he 
was right with his prediction and we are still observing the same growth 

for another 20 years than the transistors would be so small, that they 
would be fully charged by just one electron and they would shrink to 
the size of an atom. While so far micro-electronics has worked on prin- 
ciples that can be understood to a large extent using classical physics it 
is clear that transistors that are of the size of one atom must see plenty 
of quantum mechanical effects. As a matter of fact we would expect 
that quantum mechanical effects will play a significant role even ear- 
lier. This implies that we should really start to think about information 
processing and computation on the quantum mechanical level. These 
thoughts led to the birth of quantum information theory. Nevertheless 
it is not yet clear why there should be an essential difference between 
information at a classical level and information at the quantum level. 
However, this is not the case. 

A very important insight that people had is the observation that 
information should not be regarded as an isolated purely mathematical 
concept! Why is that so? You may try to define information as an 
abstract concept but you should never forget that information needs to 


be represented and transmitted. Both of these processes are physical. 
An example for the storage of information are my printed lecture notes 
which use ink on paper. Even in you brain information is stored not 
in an immaterial form, but rather in the form of synaptic connections 
between your brain cells. A computer, finally, stores information in 
transistors that are either charged or uncharged. Likewise information 
transmission requires physical objects. Talking to you means that I am 
using sound waves (described by classical physics) to send information 
to you. Television signals going through a cable represent information 
transfer using a physical system. In a computer, finally, the information 
that is stored in the transistors is transported by small currents. 

So, clearly information and physics are not two separate concepts 
but should be considered simultaneously. Our everyday experience is 
of course mainly dominated by classical physics, and therefore it is not 
surprising that our normal perception of information is governed by 
classical physics. However, the question remains, what will happen 
when we try to unify information theory with quantum physics. What 
new effects can appear? Can we do useful things with this new theory? 
These are all important questions but above all I think that it is good 
fun to learn something new about nature. 


5.1 What is information? Bits and all 

5.2 From classical information to quan- 
tum information. 

5.3 Distinguishing quantum states and the 
no-cloning theorem. 

5.4 Quantum entanglement: From qubits 
to ebits. 

5.5 Quantum state teleportation. 

5.6 Quantum dense coding. 

5.7 Local manipulation of quantum states. 

5.8 Quantum cyptography 

5.9 Quantum computation 

Quantum Bits 

Before I really start, I will have to introduce a very basic notion. The 
first one is the generalization of the classical bit. In classical information 
theory information is usually represented in the form of classical bits, 
i.e. and 1. These two values may for example be represented as an 
uncharged transistor ('0') and a fully charged transistor (T'), see Fig 

Note that a charged transistor easily holds 10 8 electrons. Therefore 
it doesn't make much difference whether a few thousand electrons are 



Figure 5.2: A transistor can represent two distinct logical values. An 
uncharged transistor represents '0' and a fully charged transistor rep- 
resent '1'. The amount of charge may vary a little bit without causing 
an error. 

missing. A transistor charged with 10 8 electrons and one with 0.999 -10 s 
electrons both represent the logical value 1. 

This situation changes, when we consider either very small tran- 
sistors or in the extreme case atoms. Imagine an atom which stores 
the numbers '0' and T' in its internal states. For example an electron 
in the ground state represents the value '0', while an electron in the 
excited state represents the value '1' (see Fig. 5.3). In the following 
we will disregard all the other energy levels and idealize the system as 
a two level system. Such a quantum mechanical two level system will 
from now on be called a quantum bit or shortly a qubit. So far this 
is just the same situation as in the classical case of a transistor. How- 
ever, there are two differences. Firstly the atomic system will be much 
more sensitive to perturbations, because now it makes a big difference 
whether there is one electron or no electron. Secondly, and probably 
more importantly, in quantum mechanics we have the superposition 
principle. Therefore we do not only have the possibilities '0' and '1' 
represented by the two quantum states |0) and |1), but we may also 
have coherent superpositions between different values (Fig 5.3), i.e. 

M = a|0) + 6|1) (5.1) 

Therefore we have the ability to represent simultaneously two values 
in a single quantum bit. But it comes even better. Imagine that you 
are holding four qubits. Then they can be in a state that is a coherent 


Figure 5.3: 

superposition of 16 different states, each representing binary strings. 

|^) = I (|0000) + |0001) + |0010) + |0011) 

+|0100) + |0101) + |0110) + |0111) 
+|1000) + |1001) + |1010) + I 1011) 
+ |1100) + |1101) + 1 1110) + I 111 1>) (5.2) 

Evidently a collection of n qubits can be in a state that is a coherent 
superposition of 2 n different quantum states, each of which represents 
a number in binary notation. If we apply a unitary transformation o 
such a state, we therefore manipulate 2 n binary numbers simultane- 
ously! This represents a massive parallelism in our computation which 
is responsible for the fact that a quantum mechanical system can solve 
certain problems exponentially faster than any classical system can do. 
However, it is very difficult to build such a quantum system in a way 
that we can control it very precisely and that it is insensitive to noise 
at the same time. This has is quite difficult, and there are no real 
quantum computers around present. 

If I have enough time, I am going to explain the idea of a quantum 
computer in more detail, but first I would like to present some effects 
and applications of quantum entanglement which have actually been 
realized in experiments. 

5.10 Entanglement and Bell inequalities 

In the previous section we have seen that the superposition principle 
is one source for the new effects that can arise in quantum informa- 
tion theory. However, the superposition principle alone exists also in 
classical physics, e.g. in sound waves or electro-magnetic waves. What 



is completely absent from classical physics is the notion of entangle- 
ment (from the German word Verschrankung) of which you have heard 
about earlier in this lecture. Entanglement was first investigated by 
Schrodinger in his Gedanken experiments using cats. Later Einstein, 
Podolsky and Rosen proposed a famous Gedanken experiment which 
they used to criticise quantum mechanics. The notion of entanglement 
arises from the superposition principle together with the tensor product 
structure of the quantum mechanical Hilbert space. States of the form 

are called product states. Such states are called disentangled as the 
outcome of a measurement on the first system is independent of the 
outcome of a measurement on the other particle. On the other hand 
states of the form 

are very strongly correlated. If a measurement on the first system shows 
that the system is in state | |) (| f)) then we immediately know that 
the second system must be in state | j) (| f)) too. If this would be 
all that can be said about the correlations in the state Eq. (5.4), then 
it would not be justified to call this states entangled simply because 
we can produce the same correlations also in a purely classical setting. 
Imagine for example, that we have two coins and that they are always 
prepared in a way, such that either both of them show heads or both 
of them show tails. Then by looking at only one of them, we know 
whether the other one shows head or tails. The correlations in the state 
Eq. (5.4), however, have much more complicated properties. These new 
properties are due to the fact, that in quantum mechanics we can make 
measurements in bases other than the {| |), | |)} basis. Any basis of 
the form {a\ |) + b\ 1),b*\ j) — a*\ |)} can also be used. This makes 
the structure of the quantum mechanical state Eq. (5.4) much richer 
than that of the example of the two coins. 

A famous example in which these new correlations manifest them- 
selves is that of the Bell inequalities. In the rest of this section I am go- 
ing to explain to you some of the ideas behind the Bell inequalities and 




their significance. When Bell started to think about the foundations 
of quantum mechanics, the work that later led to the Bell inequalities, 
he was interested in one particular problem. Since the discovery of 
quantum mechanics, physicists have been worried about the fact that 
quantum mechanical measurements lead to random measurement out- 
comes. All that quantum mechanics predicts are the probabilities for 
the different possible measurement outcomes. This is quite substan- 
tially different from everything that we know from classical physics, 
where the measurement outcomes are not random if we have complete 
knowledge of all system variables. Only incomplete knowledge of the 
system can lead to random measurement outcomes. Does that mean 
that quantum mechanics is an incomplete description of nature? Are 
there hidden variables, that we cannot observe directly, but which de- 
termine the outcome of our measurements? In a book on the mathe- 
matical foundations of quantum mechanics John von Neumann had ac- 
tually presented a proof that such theories cannot exist. Unfortunately 
the proof is wrong as von Neumann had made a wrong assumption. 
The question whether there are hidden variable theories that repro- 
duce all quantum mechanical predictions was therefore still open. John 
Bell finally re-investigated the experiment that has originally been pro- 
posed by Einstein, Podolsky and Rosen and he finally succeeded in 
showing that there is a physically observable quantity for which local 
hidden variable theories and quantum mechanics give different predic- 
tions. This was an immense step forward, because now it had become 
possible to test experimentally whether quantum mechanics with all 
its randomness is correct or whether there are hidden variable theories 
that explain the randomness in the measurement results by our incom- 
plete knowledge. In the following I am going to show you a derivation 
of Bell's inequalities. 

To do this I now need to define quantitatively what I mean by corre- 
lations. The state Eq. (5.4) describes the state of two spin-| particles. 
Assume that I measure the orientation of the first spin along direction 
| a) and that of the second spin along direction b. Both measurement 
can only have one of two outcomes, either the spin is parallel or anti- 
parallel to the direction along which it has been measured. If it is 
parallel then I assign the value a = 1 (b = 1) to the measurement out- 
come, if it is anti-parallel, then I assign the value a — — 1 (b — — 1) 



to the measurement outcome. If we repeat the measurement N times, 
each time preparing the original state Eq. (5.4) and then performing 
the measurement, then the correlation between the two measurements 
is defined as 

1 N 

C(a, b) = lim — a nK ■ (5.5) 

N^oo iv — i 

Now I want to show that correlations of the form Eq. (5.5) satisfy 
Bells inequalities, given two assumptions are being made 

1. We impose locality, i.e. a measurement on the first particle has 
no effect on the second side when the measurements are at space- 
like locations (no signal can travel after the measurement on the 
first particle from there to the second particle before the mea- 
surement on the second particle has been carried out). This leads 
to probabilities that are just products of probabilities for a mea- 
surement outcome for the first particle side and the probability 
for an measurement outcome on the second side. 

2. There are hidden variables that we cannot access directly but 
which influence the probabilities that are observed. This implies 
that all probabilities are of the form P A (a, A) and P B (b, A) where 
A describes the hidden variables. 

Under these assumptions I want to prove that for measurements 

— * — * 

along the four directions a, a', b and b' we find the Bell inequality 

\C(a,b) + C{a,b') + C{a',b) -C{a',b')\ < 2 . (5.6) 
Proof: To see this we use the fact that for all a n , a' n , b n , b' n G [—1,1] 

\a n (b n + b' n ) + a' n (b n -b' n )\ <2 . (5.7) 

Now we find 

C(a, b) = J d\p(X) [P A (+, X)P B (+, A) + P A (-, X)P B (- A) 
-Pa(+, X)Pb(-, A) - Pa(~, X)P b (+, A)] 
= j dXp(X) [P A (+, A) - P A (- A)] [P B (+, A) - P B (- A)] 

= jQ A (a,X)Q B (b,X)p(X)dX (5.8) 


and therefore 

\C{a, b) + C{a, b') + C(ct, b) - C{a, b) \ < 

J \QA{a,X)QB(b,X) + QA(a,X)QB(b',X) 

+Q A (a', X)Q B (b, A) - Q A (a', X)Q B (b', X)\p(X)dX 
< 2 

This finishes the proof. 

Therefore if we are able to explain quantum mechanics by a local 
hidden variable theory, then Eq. (5.6) has to be satisfied. 

Now let us calculate what quantum mechanics is telling us about 
the left hand side of Eq. (5.6). Quantum mechanically the correlation 
is given by 

C(a,b) = (ip\(adr) ® (adr)\ip) , (5.9) 

where a = a x e x + a y e y + a z e z with the Pauli operators &i. Now we can 
express the correlation in terms of the angle d a b between the vectors a 
and b. We find (this is an exercise for you) 

C(a,b) = -cos9 ab . (5.10) 

Now we make a particular choice for the four vectors a, a', b and b'. We 

— * ^ — * 

chose a and b parallel and a' and b' such that all four vectors lie in one 
plane. Finally we chose the angles Q a y = 6 a ib = <fi. All this is shown in 
Fig. 5.4. Inserting this choice in the left hand side of Eq. (5.6), then 
we find 

|1 + 2cos0-cos20| < 2 . (5.11) 

Plotting the function on the left hand side of Eq. (5.11) in Fig. 5.5 
we see that the inequality is actually violated for quite a wide range of 
values of <fi. The maximum value for the left hand side of Eq. (5.11) is 
given by 2.5 and is assumed for the value <fi = it/ '3. 

Of course now the big question is whether Bell's inequalities are 
violated or not. Experiments testing the validity of the Bell inequalities 
have been carried out e.g. in Paris in 1982. The idea was very simple, 
but of course it was quite difficult to actually do the experiment. A 
central source produces photons which are in the singlet state 

ir> = 4(1 t>ii>- t» • (5.12) 



Figure 5.4: The relative orientation of the four directions a, a', b and b' 
along which the spins are measured. 

One can derive identical Bell inequalities for such a state and this state 
was chosen, because it can be produced fairly easily. The way it is 
done, is via the decay of an atom in such a way that it always emits 
two photons and that the total change in angular momentum in the 
atom is zero. Then the two photons are necessarily in a spin-0 state, 
i.e. a singlet state. The two photons were then flying away in opposite 
directions towards to measurement apparatuses. These apparatuses 
were then measuring the polarization state of the two photons along 
four possible directions that have been chosen to give maximal violation 
of Bell inequalities. After each decay for each apparatus a direction 
was chosen randomly and independently from the other side. Finally 
the measurement result was noted down. This experiment found a 
value for the correlations of about 2.7 which is reasonably close to the 
quantum mechanically predicted value of 2.82. More recently more 
precise experiments have been carried out and the violation of Bells 
inequalities has been clearly demonstrated. 


Figure 5.5: The right hand side of Eq. (5.11) is plotted. You can clearly 
see that it can exceed the value of 2 and achieves a maximum of 2.5. 

5.11 Quantum State Teleportation 

The procedure we will analyse is called quantum teleportation and can 
be understood as follows. The naive idea of teleportation involves a 
protocol whereby an object positioned at a place A and time t first 
"dematerializes" and then reappears at a distant place B at some later 
time t + T. Quantum teleportation implies that we wish to apply this 
procedure to a quantum object. However, a genuine quantum telepor- 
tation differs from this idea, because we are not teleporting the whole 
object but just its state from particle A to particle B. As quantum 
particles are indistinguishable anyway, this amounts to 'real' telepor- 
tation. One way of performing teleportation (and certainly the way 
portrayed in various science fiction movies, e.g. The Fly) is first to 



learn all the properties of that object (thereby possibly destroying it). 
We then send this information as a classical string of data to B where 
another object with the same properties is re-created. One problem 
with this picture is that, if we have a single quantum system in an un- 
known state, we cannot determine its state completely because of the 
uncertainty principle. More precisely, we need an infinite ensemble of 
identically prepared quantum systems to be able completely to deter- 
mine its quantum state. So it would seem that the laws of quantum 
mechanics prohibit teleportation of single quantum systems. However, 
the very feature of quantum mechanics that leads to the uncertainty 
principle (the superposition principle) also allows the existence of en- 
tangled states. These entangled states will provide a form of quantum 
channel to conduct a teleportation protocol. It will turn out that there 
is no need to learn the state of the system in order to teleport it. On the 
other hand, there is a need to send some classical information from A to 
B, but part of the information also travels down an entangled channel. 
This then provides a way of distinguishing quantum and classical cor- 
relations, which we said was at the heart of quantifying entanglement. 
After the teleportation is completed, the original state of the particle 
at A is destroyed (although the particle itself remains intact) and so 
is the entanglement in the quantum channel. These two features are 
direct consequences of fundamental laws in information processing. I 
cannot explain these here as I do not have enough time, but if you are 
interested you should have a look at the article M.B. Plenio and V. 
Vedral, Contemp. Physics 39, 431 (1998) which has been written for 
final year students and first year PhD students. 

5.12 A basic description of teleportation 

Let us begin by describing quantum teleportation in the form originally 
proposed by Bennett, Brassard, Crepeau, Jozsa, Peres, and Wootters 
in 1993. Suppose that Alice and Bob, who are distant from each other, 
wish to implement a teleportation procedure. Initially they need to 
share a maximally entangled pair of quantum mechanical two level 
systems. Unlike the classical bit, a qubit can be in a superposition of 
its basis states, like = a\0) + b\l). This means that if Alice and 


Bob both have one qubit each then the joint state may for example be 
\*ab) = (|0a)|0b) + |1a)|1b))A/2 , (5.13) 

where the first ket (with subscript A) belongs to Alice and second (with 
subscript B) to Bob. This state is entangled meaning, that it cannot be 
written as a product of the individual states (like e.g. 1 00) ) . Note that 
this state is different from a statistical mixture ( 1 00) (00 1 + |ll)(ll|)/2 
which is the most correlated state allowed by classical physics. 

Now suppose that Alice receives a qubit in a state which is unknown 
to her (let us label it |<&) = a\0) + and she has to teleport it to 
Bob. The state has to be unknown to her because otherwise she can 
just phone Bob up and tell him all the details of the state, and he 
can then recreate it on a particle that he possesses. If Alice does not 
know the state, then she cannot measure it to obtain all the necessary 
information to specify it. Therefore she has to resort to using the state 
\^ab) that she shares with Bob. To see what she has to do, we write 
out the total state of all three qubits 



\$)\*ab) = H0> + 6|1»(|00> + \U))/V2 


However, the above state can be written in the following convenient way 
(here we are only rewriting the above expression in a different basis, 
and there is no physical process taking place in between) 


(a|000) + a|011) + 6|100) + 6|111 

|$ + )(a|0) + 6|l)) + |$-)(a|0)-6|l)) 

+ |^ + )(a|l)+6|0)) + |^-)(a|l)-6|0)) , (5.15) 



(|00) + |11»/V2 
(|00) - |ll»/>/2 
(|01) + \W))/V2 



form an ortho-normal basis of Alice's two qubits (remember that the 
first two qubits belong to Alice and the last qubit belongs to Bob). 



The above basis is frequently called the Bell basis. This is a very useful 
way of writing the state of Alice's two qubits and Bob's single qubit be- 
cause it displays a high degree of correlations between Alice's and Bob's 
parts: to every state of Alice's two qubits (i.e. |<& + ), |$~), \^ + ), l^ - )) 
corresponds a state of Bob's qubit. In addition the state of Bob's qubit 
in all four cases looks very much like the original qubit that Alice has 
to teleport to Bob. It is now straightforward to see how to proceed 
with the teleportation protocol: 

1. Upon receiving the unknown qubit in state |$) Alice performs 
projective measurements on her two qubits in the Bell basis. This 
means that she will obtain one of the four Bell states randomly, 
and with equal probability. 

2. Suppose Alice obtains the state \^ + ). Then the state of all three 
qubits (Alice + Bob) collapses to the following state 

|^ + )(a|l) +6|0» . (5.20) 

(the last qubit belongs to Bob as usual). Alice now has to com- 
municate the result of her measurement to Bob (over the phone, 
for example). The point of this communication is to inform Bob 
how the state of his qubit now differs from the state of the qubit 
Alice was holding previously. 

3. Now Bob knows exactly what to do in order to complete the 
teleportation. He has to apply a unitary transformation on his 
qubit which simulates a logical NOT operation: |0) — > |1) and 
|1) — > |0). He thereby transforms the state of his qubit into 
the state a\0) +b\l), which is precisely the state that Alice had 
to teleport to him initially. This completes the protocol. It is 
easy to see that if Alice obtained some other Bell state then Bob 
would have to apply some other simple operation to complete 
teleportation. We leave it to the reader to work out the other two 
operations (note that if Alice obtained |$ + ) he would not have 
to do anything). If |0) and |1) are written in their vector form 
then the operations that Bob has to perform can be represented 
by the Pauli spin matrices, as depicted in Fig. 5.6. 



(a) © (=) > (=) (a|0)+/3|l))(|00) + |ll))/V2 

(b) |@ (^w^f*~^(E) 



+ i|*-)(a|0>-/9|l» 
+ i|*+)(a|l>+/3|0» 
+ i|*->(a|l>-/9|0» 



Alice finds I $ + ) 

Alice finds | $ ) 

Alice) finds] ^ + ) 

Alice findsl 
i 1 ' 

>° Bob does nothing 

> x Bob performs a z 

> 2 Bob performs a x 

> 3 Bob performs a z a x 

(d) © © 

Figure 5.6: 

Figure 5.6: The basic steps of quantum state teleportation. Alice 
and Bob are spatially separated, Alice on the left of the dashed line, 
Bob on the right, (a) Alice and Bob share a maximally entangled pair 
of particles in the state ( 1 00) + 111))/ y/2. Alice wants to teleport the un- 
known state to Bob. (b) The total state of the three particles that 
Alice and Bob are holding is rewritten in the Bell basis Eqs. (5.16-5.19) 
for the two particles Alice is holding. Alice performs a measurement 
that projects the state of her two particles onto one of the four Bell 
states, (c) She transmits the result encoded in the numbers 0, 1, 2, 3 
to Bob, who performs a unitary transformation 1, a z , a x , o z a x that de- 
pends only on the measurement result that Alice obtained but not on 
the state j^)! (d) After Bob has applied the appropriate unitary op- 
eration on his particle he can be sure that he is now holding the state 
that Alice was holding in (a). 



An important fact to observe in the above protocol is that all the 
operations (Alice's measurements and Bob's unitary transformations) 
are local in nature. This means that there is never any need to perform 
a (global) transformation or measurement on all three qubits simulta- 
neously, which is what allows us to call the above protocol a genuine 
teleportation. It is also important that the operations that Bob per- 
forms are independent of the state that Alice tries to teleport to Bob. 
Note also that the classical communication from Alice to Bob in step 2 
above is crucial because otherwise the protocol would be impossible to 
execute (there is a deeper reason for this: if we could perform telepor- 
tation without classical communication then Alice could send messages 
to Bob faster than the speed of light, remember that I explained this 
in a previous lecture. 

Important to observe is also the fact that the initial state to be tele- 
ported is at the end destroyed, i.e it becomes maximally mixed, of the 
form (|0)(0| + |l)(l|)/2. This has to happen since otherwise we would 
end up with two qubits in the same state at the end of teleportation 
(one with Alice and the other one with Bob). So, effectively, we would 
clone an unknown quantum state, which is impossible by the laws of 
quantum mechanics (this is the no-cloning theorem of Wootters and 
Zurek). I will explain this in more detail later. 

For those of you who are interested in quantum information theory, 
here are some references for further reading: 

L. Hardy, Cont. Physics 39, 415 (1998) 
M.B. Plenio and V. Vedral, Cont. Physics 39, 431 (1998) 
J. Preskill, http:/ /www.